-0.000 000 000 742 147 676 646 709 46 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 709 46(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
-0.000 000 000 742 147 676 646 709 46(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:

|-0.000 000 000 742 147 676 646 709 46| = 0.000 000 000 742 147 676 646 709 46


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: 0.000 000 000 742 147 676 646 709 46.

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) 0.000 000 000 742 147 676 646 709 46 × 2 = 0 + 0.000 000 001 484 295 353 293 418 92;
  • 2) 0.000 000 001 484 295 353 293 418 92 × 2 = 0 + 0.000 000 002 968 590 706 586 837 84;
  • 3) 0.000 000 002 968 590 706 586 837 84 × 2 = 0 + 0.000 000 005 937 181 413 173 675 68;
  • 4) 0.000 000 005 937 181 413 173 675 68 × 2 = 0 + 0.000 000 011 874 362 826 347 351 36;
  • 5) 0.000 000 011 874 362 826 347 351 36 × 2 = 0 + 0.000 000 023 748 725 652 694 702 72;
  • 6) 0.000 000 023 748 725 652 694 702 72 × 2 = 0 + 0.000 000 047 497 451 305 389 405 44;
  • 7) 0.000 000 047 497 451 305 389 405 44 × 2 = 0 + 0.000 000 094 994 902 610 778 810 88;
  • 8) 0.000 000 094 994 902 610 778 810 88 × 2 = 0 + 0.000 000 189 989 805 221 557 621 76;
  • 9) 0.000 000 189 989 805 221 557 621 76 × 2 = 0 + 0.000 000 379 979 610 443 115 243 52;
  • 10) 0.000 000 379 979 610 443 115 243 52 × 2 = 0 + 0.000 000 759 959 220 886 230 487 04;
  • 11) 0.000 000 759 959 220 886 230 487 04 × 2 = 0 + 0.000 001 519 918 441 772 460 974 08;
  • 12) 0.000 001 519 918 441 772 460 974 08 × 2 = 0 + 0.000 003 039 836 883 544 921 948 16;
  • 13) 0.000 003 039 836 883 544 921 948 16 × 2 = 0 + 0.000 006 079 673 767 089 843 896 32;
  • 14) 0.000 006 079 673 767 089 843 896 32 × 2 = 0 + 0.000 012 159 347 534 179 687 792 64;
  • 15) 0.000 012 159 347 534 179 687 792 64 × 2 = 0 + 0.000 024 318 695 068 359 375 585 28;
  • 16) 0.000 024 318 695 068 359 375 585 28 × 2 = 0 + 0.000 048 637 390 136 718 751 170 56;
  • 17) 0.000 048 637 390 136 718 751 170 56 × 2 = 0 + 0.000 097 274 780 273 437 502 341 12;
  • 18) 0.000 097 274 780 273 437 502 341 12 × 2 = 0 + 0.000 194 549 560 546 875 004 682 24;
  • 19) 0.000 194 549 560 546 875 004 682 24 × 2 = 0 + 0.000 389 099 121 093 750 009 364 48;
  • 20) 0.000 389 099 121 093 750 009 364 48 × 2 = 0 + 0.000 778 198 242 187 500 018 728 96;
  • 21) 0.000 778 198 242 187 500 018 728 96 × 2 = 0 + 0.001 556 396 484 375 000 037 457 92;
  • 22) 0.001 556 396 484 375 000 037 457 92 × 2 = 0 + 0.003 112 792 968 750 000 074 915 84;
  • 23) 0.003 112 792 968 750 000 074 915 84 × 2 = 0 + 0.006 225 585 937 500 000 149 831 68;
  • 24) 0.006 225 585 937 500 000 149 831 68 × 2 = 0 + 0.012 451 171 875 000 000 299 663 36;
  • 25) 0.012 451 171 875 000 000 299 663 36 × 2 = 0 + 0.024 902 343 750 000 000 599 326 72;
  • 26) 0.024 902 343 750 000 000 599 326 72 × 2 = 0 + 0.049 804 687 500 000 001 198 653 44;
  • 27) 0.049 804 687 500 000 001 198 653 44 × 2 = 0 + 0.099 609 375 000 000 002 397 306 88;
  • 28) 0.099 609 375 000 000 002 397 306 88 × 2 = 0 + 0.199 218 750 000 000 004 794 613 76;
  • 29) 0.199 218 750 000 000 004 794 613 76 × 2 = 0 + 0.398 437 500 000 000 009 589 227 52;
  • 30) 0.398 437 500 000 000 009 589 227 52 × 2 = 0 + 0.796 875 000 000 000 019 178 455 04;
  • 31) 0.796 875 000 000 000 019 178 455 04 × 2 = 1 + 0.593 750 000 000 000 038 356 910 08;
  • 32) 0.593 750 000 000 000 038 356 910 08 × 2 = 1 + 0.187 500 000 000 000 076 713 820 16;
  • 33) 0.187 500 000 000 000 076 713 820 16 × 2 = 0 + 0.375 000 000 000 000 153 427 640 32;
  • 34) 0.375 000 000 000 000 153 427 640 32 × 2 = 0 + 0.750 000 000 000 000 306 855 280 64;
  • 35) 0.750 000 000 000 000 306 855 280 64 × 2 = 1 + 0.500 000 000 000 000 613 710 561 28;
  • 36) 0.500 000 000 000 000 613 710 561 28 × 2 = 1 + 0.000 000 000 000 001 227 421 122 56;
  • 37) 0.000 000 000 000 001 227 421 122 56 × 2 = 0 + 0.000 000 000 000 002 454 842 245 12;
  • 38) 0.000 000 000 000 002 454 842 245 12 × 2 = 0 + 0.000 000 000 000 004 909 684 490 24;
  • 39) 0.000 000 000 000 004 909 684 490 24 × 2 = 0 + 0.000 000 000 000 009 819 368 980 48;
  • 40) 0.000 000 000 000 009 819 368 980 48 × 2 = 0 + 0.000 000 000 000 019 638 737 960 96;
  • 41) 0.000 000 000 000 019 638 737 960 96 × 2 = 0 + 0.000 000 000 000 039 277 475 921 92;
  • 42) 0.000 000 000 000 039 277 475 921 92 × 2 = 0 + 0.000 000 000 000 078 554 951 843 84;
  • 43) 0.000 000 000 000 078 554 951 843 84 × 2 = 0 + 0.000 000 000 000 157 109 903 687 68;
  • 44) 0.000 000 000 000 157 109 903 687 68 × 2 = 0 + 0.000 000 000 000 314 219 807 375 36;
  • 45) 0.000 000 000 000 314 219 807 375 36 × 2 = 0 + 0.000 000 000 000 628 439 614 750 72;
  • 46) 0.000 000 000 000 628 439 614 750 72 × 2 = 0 + 0.000 000 000 001 256 879 229 501 44;
  • 47) 0.000 000 000 001 256 879 229 501 44 × 2 = 0 + 0.000 000 000 002 513 758 459 002 88;
  • 48) 0.000 000 000 002 513 758 459 002 88 × 2 = 0 + 0.000 000 000 005 027 516 918 005 76;
  • 49) 0.000 000 000 005 027 516 918 005 76 × 2 = 0 + 0.000 000 000 010 055 033 836 011 52;
  • 50) 0.000 000 000 010 055 033 836 011 52 × 2 = 0 + 0.000 000 000 020 110 067 672 023 04;
  • 51) 0.000 000 000 020 110 067 672 023 04 × 2 = 0 + 0.000 000 000 040 220 135 344 046 08;
  • 52) 0.000 000 000 040 220 135 344 046 08 × 2 = 0 + 0.000 000 000 080 440 270 688 092 16;
  • 53) 0.000 000 000 080 440 270 688 092 16 × 2 = 0 + 0.000 000 000 160 880 541 376 184 32;
  • 54) 0.000 000 000 160 880 541 376 184 32 × 2 = 0 + 0.000 000 000 321 761 082 752 368 64;

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:


0.000 000 000 742 147 676 646 709 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 709 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 709 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 1000 0000 0000 0000 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 0000(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. 100 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 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) =
0110 0000

Mantissa (23 bits) =
100 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 709 46 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 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