-0.000 000 000 742 147 676 635 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 635(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
-0.000 000 000 742 147 676 635(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:

|-0.000 000 000 742 147 676 635| = 0.000 000 000 742 147 676 635


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: 0.000 000 000 742 147 676 635.

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) 0.000 000 000 742 147 676 635 × 2 = 0 + 0.000 000 001 484 295 353 27;
  • 2) 0.000 000 001 484 295 353 27 × 2 = 0 + 0.000 000 002 968 590 706 54;
  • 3) 0.000 000 002 968 590 706 54 × 2 = 0 + 0.000 000 005 937 181 413 08;
  • 4) 0.000 000 005 937 181 413 08 × 2 = 0 + 0.000 000 011 874 362 826 16;
  • 5) 0.000 000 011 874 362 826 16 × 2 = 0 + 0.000 000 023 748 725 652 32;
  • 6) 0.000 000 023 748 725 652 32 × 2 = 0 + 0.000 000 047 497 451 304 64;
  • 7) 0.000 000 047 497 451 304 64 × 2 = 0 + 0.000 000 094 994 902 609 28;
  • 8) 0.000 000 094 994 902 609 28 × 2 = 0 + 0.000 000 189 989 805 218 56;
  • 9) 0.000 000 189 989 805 218 56 × 2 = 0 + 0.000 000 379 979 610 437 12;
  • 10) 0.000 000 379 979 610 437 12 × 2 = 0 + 0.000 000 759 959 220 874 24;
  • 11) 0.000 000 759 959 220 874 24 × 2 = 0 + 0.000 001 519 918 441 748 48;
  • 12) 0.000 001 519 918 441 748 48 × 2 = 0 + 0.000 003 039 836 883 496 96;
  • 13) 0.000 003 039 836 883 496 96 × 2 = 0 + 0.000 006 079 673 766 993 92;
  • 14) 0.000 006 079 673 766 993 92 × 2 = 0 + 0.000 012 159 347 533 987 84;
  • 15) 0.000 012 159 347 533 987 84 × 2 = 0 + 0.000 024 318 695 067 975 68;
  • 16) 0.000 024 318 695 067 975 68 × 2 = 0 + 0.000 048 637 390 135 951 36;
  • 17) 0.000 048 637 390 135 951 36 × 2 = 0 + 0.000 097 274 780 271 902 72;
  • 18) 0.000 097 274 780 271 902 72 × 2 = 0 + 0.000 194 549 560 543 805 44;
  • 19) 0.000 194 549 560 543 805 44 × 2 = 0 + 0.000 389 099 121 087 610 88;
  • 20) 0.000 389 099 121 087 610 88 × 2 = 0 + 0.000 778 198 242 175 221 76;
  • 21) 0.000 778 198 242 175 221 76 × 2 = 0 + 0.001 556 396 484 350 443 52;
  • 22) 0.001 556 396 484 350 443 52 × 2 = 0 + 0.003 112 792 968 700 887 04;
  • 23) 0.003 112 792 968 700 887 04 × 2 = 0 + 0.006 225 585 937 401 774 08;
  • 24) 0.006 225 585 937 401 774 08 × 2 = 0 + 0.012 451 171 874 803 548 16;
  • 25) 0.012 451 171 874 803 548 16 × 2 = 0 + 0.024 902 343 749 607 096 32;
  • 26) 0.024 902 343 749 607 096 32 × 2 = 0 + 0.049 804 687 499 214 192 64;
  • 27) 0.049 804 687 499 214 192 64 × 2 = 0 + 0.099 609 374 998 428 385 28;
  • 28) 0.099 609 374 998 428 385 28 × 2 = 0 + 0.199 218 749 996 856 770 56;
  • 29) 0.199 218 749 996 856 770 56 × 2 = 0 + 0.398 437 499 993 713 541 12;
  • 30) 0.398 437 499 993 713 541 12 × 2 = 0 + 0.796 874 999 987 427 082 24;
  • 31) 0.796 874 999 987 427 082 24 × 2 = 1 + 0.593 749 999 974 854 164 48;
  • 32) 0.593 749 999 974 854 164 48 × 2 = 1 + 0.187 499 999 949 708 328 96;
  • 33) 0.187 499 999 949 708 328 96 × 2 = 0 + 0.374 999 999 899 416 657 92;
  • 34) 0.374 999 999 899 416 657 92 × 2 = 0 + 0.749 999 999 798 833 315 84;
  • 35) 0.749 999 999 798 833 315 84 × 2 = 1 + 0.499 999 999 597 666 631 68;
  • 36) 0.499 999 999 597 666 631 68 × 2 = 0 + 0.999 999 999 195 333 263 36;
  • 37) 0.999 999 999 195 333 263 36 × 2 = 1 + 0.999 999 998 390 666 526 72;
  • 38) 0.999 999 998 390 666 526 72 × 2 = 1 + 0.999 999 996 781 333 053 44;
  • 39) 0.999 999 996 781 333 053 44 × 2 = 1 + 0.999 999 993 562 666 106 88;
  • 40) 0.999 999 993 562 666 106 88 × 2 = 1 + 0.999 999 987 125 332 213 76;
  • 41) 0.999 999 987 125 332 213 76 × 2 = 1 + 0.999 999 974 250 664 427 52;
  • 42) 0.999 999 974 250 664 427 52 × 2 = 1 + 0.999 999 948 501 328 855 04;
  • 43) 0.999 999 948 501 328 855 04 × 2 = 1 + 0.999 999 897 002 657 710 08;
  • 44) 0.999 999 897 002 657 710 08 × 2 = 1 + 0.999 999 794 005 315 420 16;
  • 45) 0.999 999 794 005 315 420 16 × 2 = 1 + 0.999 999 588 010 630 840 32;
  • 46) 0.999 999 588 010 630 840 32 × 2 = 1 + 0.999 999 176 021 261 680 64;
  • 47) 0.999 999 176 021 261 680 64 × 2 = 1 + 0.999 998 352 042 523 361 28;
  • 48) 0.999 998 352 042 523 361 28 × 2 = 1 + 0.999 996 704 085 046 722 56;
  • 49) 0.999 996 704 085 046 722 56 × 2 = 1 + 0.999 993 408 170 093 445 12;
  • 50) 0.999 993 408 170 093 445 12 × 2 = 1 + 0.999 986 816 340 186 890 24;
  • 51) 0.999 986 816 340 186 890 24 × 2 = 1 + 0.999 973 632 680 373 780 48;
  • 52) 0.999 973 632 680 373 780 48 × 2 = 1 + 0.999 947 265 360 747 560 96;
  • 53) 0.999 947 265 360 747 560 96 × 2 = 1 + 0.999 894 530 721 495 121 92;
  • 54) 0.999 894 530 721 495 121 92 × 2 = 1 + 0.999 789 061 442 990 243 84;

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:


0.000 000 000 742 147 676 635(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 635(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 635(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 635 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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