-0.000 000 000 000 000 026 72 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 026 72(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 000 026 72(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 000 026 72| = 0.000 000 000 000 000 026 72


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 026 72.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 000 026 72 × 2 = 0 + 0.000 000 000 000 000 053 44;
  • 2) 0.000 000 000 000 000 053 44 × 2 = 0 + 0.000 000 000 000 000 106 88;
  • 3) 0.000 000 000 000 000 106 88 × 2 = 0 + 0.000 000 000 000 000 213 76;
  • 4) 0.000 000 000 000 000 213 76 × 2 = 0 + 0.000 000 000 000 000 427 52;
  • 5) 0.000 000 000 000 000 427 52 × 2 = 0 + 0.000 000 000 000 000 855 04;
  • 6) 0.000 000 000 000 000 855 04 × 2 = 0 + 0.000 000 000 000 001 710 08;
  • 7) 0.000 000 000 000 001 710 08 × 2 = 0 + 0.000 000 000 000 003 420 16;
  • 8) 0.000 000 000 000 003 420 16 × 2 = 0 + 0.000 000 000 000 006 840 32;
  • 9) 0.000 000 000 000 006 840 32 × 2 = 0 + 0.000 000 000 000 013 680 64;
  • 10) 0.000 000 000 000 013 680 64 × 2 = 0 + 0.000 000 000 000 027 361 28;
  • 11) 0.000 000 000 000 027 361 28 × 2 = 0 + 0.000 000 000 000 054 722 56;
  • 12) 0.000 000 000 000 054 722 56 × 2 = 0 + 0.000 000 000 000 109 445 12;
  • 13) 0.000 000 000 000 109 445 12 × 2 = 0 + 0.000 000 000 000 218 890 24;
  • 14) 0.000 000 000 000 218 890 24 × 2 = 0 + 0.000 000 000 000 437 780 48;
  • 15) 0.000 000 000 000 437 780 48 × 2 = 0 + 0.000 000 000 000 875 560 96;
  • 16) 0.000 000 000 000 875 560 96 × 2 = 0 + 0.000 000 000 001 751 121 92;
  • 17) 0.000 000 000 001 751 121 92 × 2 = 0 + 0.000 000 000 003 502 243 84;
  • 18) 0.000 000 000 003 502 243 84 × 2 = 0 + 0.000 000 000 007 004 487 68;
  • 19) 0.000 000 000 007 004 487 68 × 2 = 0 + 0.000 000 000 014 008 975 36;
  • 20) 0.000 000 000 014 008 975 36 × 2 = 0 + 0.000 000 000 028 017 950 72;
  • 21) 0.000 000 000 028 017 950 72 × 2 = 0 + 0.000 000 000 056 035 901 44;
  • 22) 0.000 000 000 056 035 901 44 × 2 = 0 + 0.000 000 000 112 071 802 88;
  • 23) 0.000 000 000 112 071 802 88 × 2 = 0 + 0.000 000 000 224 143 605 76;
  • 24) 0.000 000 000 224 143 605 76 × 2 = 0 + 0.000 000 000 448 287 211 52;
  • 25) 0.000 000 000 448 287 211 52 × 2 = 0 + 0.000 000 000 896 574 423 04;
  • 26) 0.000 000 000 896 574 423 04 × 2 = 0 + 0.000 000 001 793 148 846 08;
  • 27) 0.000 000 001 793 148 846 08 × 2 = 0 + 0.000 000 003 586 297 692 16;
  • 28) 0.000 000 003 586 297 692 16 × 2 = 0 + 0.000 000 007 172 595 384 32;
  • 29) 0.000 000 007 172 595 384 32 × 2 = 0 + 0.000 000 014 345 190 768 64;
  • 30) 0.000 000 014 345 190 768 64 × 2 = 0 + 0.000 000 028 690 381 537 28;
  • 31) 0.000 000 028 690 381 537 28 × 2 = 0 + 0.000 000 057 380 763 074 56;
  • 32) 0.000 000 057 380 763 074 56 × 2 = 0 + 0.000 000 114 761 526 149 12;
  • 33) 0.000 000 114 761 526 149 12 × 2 = 0 + 0.000 000 229 523 052 298 24;
  • 34) 0.000 000 229 523 052 298 24 × 2 = 0 + 0.000 000 459 046 104 596 48;
  • 35) 0.000 000 459 046 104 596 48 × 2 = 0 + 0.000 000 918 092 209 192 96;
  • 36) 0.000 000 918 092 209 192 96 × 2 = 0 + 0.000 001 836 184 418 385 92;
  • 37) 0.000 001 836 184 418 385 92 × 2 = 0 + 0.000 003 672 368 836 771 84;
  • 38) 0.000 003 672 368 836 771 84 × 2 = 0 + 0.000 007 344 737 673 543 68;
  • 39) 0.000 007 344 737 673 543 68 × 2 = 0 + 0.000 014 689 475 347 087 36;
  • 40) 0.000 014 689 475 347 087 36 × 2 = 0 + 0.000 029 378 950 694 174 72;
  • 41) 0.000 029 378 950 694 174 72 × 2 = 0 + 0.000 058 757 901 388 349 44;
  • 42) 0.000 058 757 901 388 349 44 × 2 = 0 + 0.000 117 515 802 776 698 88;
  • 43) 0.000 117 515 802 776 698 88 × 2 = 0 + 0.000 235 031 605 553 397 76;
  • 44) 0.000 235 031 605 553 397 76 × 2 = 0 + 0.000 470 063 211 106 795 52;
  • 45) 0.000 470 063 211 106 795 52 × 2 = 0 + 0.000 940 126 422 213 591 04;
  • 46) 0.000 940 126 422 213 591 04 × 2 = 0 + 0.001 880 252 844 427 182 08;
  • 47) 0.001 880 252 844 427 182 08 × 2 = 0 + 0.003 760 505 688 854 364 16;
  • 48) 0.003 760 505 688 854 364 16 × 2 = 0 + 0.007 521 011 377 708 728 32;
  • 49) 0.007 521 011 377 708 728 32 × 2 = 0 + 0.015 042 022 755 417 456 64;
  • 50) 0.015 042 022 755 417 456 64 × 2 = 0 + 0.030 084 045 510 834 913 28;
  • 51) 0.030 084 045 510 834 913 28 × 2 = 0 + 0.060 168 091 021 669 826 56;
  • 52) 0.060 168 091 021 669 826 56 × 2 = 0 + 0.120 336 182 043 339 653 12;
  • 53) 0.120 336 182 043 339 653 12 × 2 = 0 + 0.240 672 364 086 679 306 24;
  • 54) 0.240 672 364 086 679 306 24 × 2 = 0 + 0.481 344 728 173 358 612 48;
  • 55) 0.481 344 728 173 358 612 48 × 2 = 0 + 0.962 689 456 346 717 224 96;
  • 56) 0.962 689 456 346 717 224 96 × 2 = 1 + 0.925 378 912 693 434 449 92;
  • 57) 0.925 378 912 693 434 449 92 × 2 = 1 + 0.850 757 825 386 868 899 84;
  • 58) 0.850 757 825 386 868 899 84 × 2 = 1 + 0.701 515 650 773 737 799 68;
  • 59) 0.701 515 650 773 737 799 68 × 2 = 1 + 0.403 031 301 547 475 599 36;
  • 60) 0.403 031 301 547 475 599 36 × 2 = 0 + 0.806 062 603 094 951 198 72;
  • 61) 0.806 062 603 094 951 198 72 × 2 = 1 + 0.612 125 206 189 902 397 44;
  • 62) 0.612 125 206 189 902 397 44 × 2 = 1 + 0.224 250 412 379 804 794 88;
  • 63) 0.224 250 412 379 804 794 88 × 2 = 0 + 0.448 500 824 759 609 589 76;
  • 64) 0.448 500 824 759 609 589 76 × 2 = 0 + 0.897 001 649 519 219 179 52;
  • 65) 0.897 001 649 519 219 179 52 × 2 = 1 + 0.794 003 299 038 438 359 04;
  • 66) 0.794 003 299 038 438 359 04 × 2 = 1 + 0.588 006 598 076 876 718 08;
  • 67) 0.588 006 598 076 876 718 08 × 2 = 1 + 0.176 013 196 153 753 436 16;
  • 68) 0.176 013 196 153 753 436 16 × 2 = 0 + 0.352 026 392 307 506 872 32;
  • 69) 0.352 026 392 307 506 872 32 × 2 = 0 + 0.704 052 784 615 013 744 64;
  • 70) 0.704 052 784 615 013 744 64 × 2 = 1 + 0.408 105 569 230 027 489 28;
  • 71) 0.408 105 569 230 027 489 28 × 2 = 0 + 0.816 211 138 460 054 978 56;
  • 72) 0.816 211 138 460 054 978 56 × 2 = 1 + 0.632 422 276 920 109 957 12;
  • 73) 0.632 422 276 920 109 957 12 × 2 = 1 + 0.264 844 553 840 219 914 24;
  • 74) 0.264 844 553 840 219 914 24 × 2 = 0 + 0.529 689 107 680 439 828 48;
  • 75) 0.529 689 107 680 439 828 48 × 2 = 1 + 0.059 378 215 360 879 656 96;
  • 76) 0.059 378 215 360 879 656 96 × 2 = 0 + 0.118 756 430 721 759 313 92;
  • 77) 0.118 756 430 721 759 313 92 × 2 = 0 + 0.237 512 861 443 518 627 84;
  • 78) 0.237 512 861 443 518 627 84 × 2 = 0 + 0.475 025 722 887 037 255 68;
  • 79) 0.475 025 722 887 037 255 68 × 2 = 0 + 0.950 051 445 774 074 511 36;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 000 026 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1110 1100 1110 0101 1010 000(2)

6. Positive number before normalization:

0.000 000 000 000 000 026 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1110 1100 1110 0101 1010 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 56 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 026 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1110 1100 1110 0101 1010 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1110 1100 1110 0101 1010 000(2) × 20 =

1.1110 1100 1110 0101 1010 000(2) × 2-56


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -56

Mantissa (not normalized):
1.1110 1100 1110 0101 1010 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-56 + 2(8-1) - 1 =

(-56 + 127)(10) =

71(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 71 ÷ 2 = 35 + 1;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

71(10) =

0100 0111(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 111 0110 0111 0010 1101 0000 =

111 0110 0111 0010 1101 0000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0100 0111

Mantissa (23 bits) =
111 0110 0111 0010 1101 0000


Decimal number -0.000 000 000 000 000 026 72 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0100 0111 - 111 0110 0111 0010 1101 0000

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111