-0.000 000 000 742 147 675 32 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 32(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
-0.000 000 000 742 147 675 32(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:

|-0.000 000 000 742 147 675 32| = 0.000 000 000 742 147 675 32


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: 0.000 000 000 742 147 675 32.

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) 0.000 000 000 742 147 675 32 × 2 = 0 + 0.000 000 001 484 295 350 64;
  • 2) 0.000 000 001 484 295 350 64 × 2 = 0 + 0.000 000 002 968 590 701 28;
  • 3) 0.000 000 002 968 590 701 28 × 2 = 0 + 0.000 000 005 937 181 402 56;
  • 4) 0.000 000 005 937 181 402 56 × 2 = 0 + 0.000 000 011 874 362 805 12;
  • 5) 0.000 000 011 874 362 805 12 × 2 = 0 + 0.000 000 023 748 725 610 24;
  • 6) 0.000 000 023 748 725 610 24 × 2 = 0 + 0.000 000 047 497 451 220 48;
  • 7) 0.000 000 047 497 451 220 48 × 2 = 0 + 0.000 000 094 994 902 440 96;
  • 8) 0.000 000 094 994 902 440 96 × 2 = 0 + 0.000 000 189 989 804 881 92;
  • 9) 0.000 000 189 989 804 881 92 × 2 = 0 + 0.000 000 379 979 609 763 84;
  • 10) 0.000 000 379 979 609 763 84 × 2 = 0 + 0.000 000 759 959 219 527 68;
  • 11) 0.000 000 759 959 219 527 68 × 2 = 0 + 0.000 001 519 918 439 055 36;
  • 12) 0.000 001 519 918 439 055 36 × 2 = 0 + 0.000 003 039 836 878 110 72;
  • 13) 0.000 003 039 836 878 110 72 × 2 = 0 + 0.000 006 079 673 756 221 44;
  • 14) 0.000 006 079 673 756 221 44 × 2 = 0 + 0.000 012 159 347 512 442 88;
  • 15) 0.000 012 159 347 512 442 88 × 2 = 0 + 0.000 024 318 695 024 885 76;
  • 16) 0.000 024 318 695 024 885 76 × 2 = 0 + 0.000 048 637 390 049 771 52;
  • 17) 0.000 048 637 390 049 771 52 × 2 = 0 + 0.000 097 274 780 099 543 04;
  • 18) 0.000 097 274 780 099 543 04 × 2 = 0 + 0.000 194 549 560 199 086 08;
  • 19) 0.000 194 549 560 199 086 08 × 2 = 0 + 0.000 389 099 120 398 172 16;
  • 20) 0.000 389 099 120 398 172 16 × 2 = 0 + 0.000 778 198 240 796 344 32;
  • 21) 0.000 778 198 240 796 344 32 × 2 = 0 + 0.001 556 396 481 592 688 64;
  • 22) 0.001 556 396 481 592 688 64 × 2 = 0 + 0.003 112 792 963 185 377 28;
  • 23) 0.003 112 792 963 185 377 28 × 2 = 0 + 0.006 225 585 926 370 754 56;
  • 24) 0.006 225 585 926 370 754 56 × 2 = 0 + 0.012 451 171 852 741 509 12;
  • 25) 0.012 451 171 852 741 509 12 × 2 = 0 + 0.024 902 343 705 483 018 24;
  • 26) 0.024 902 343 705 483 018 24 × 2 = 0 + 0.049 804 687 410 966 036 48;
  • 27) 0.049 804 687 410 966 036 48 × 2 = 0 + 0.099 609 374 821 932 072 96;
  • 28) 0.099 609 374 821 932 072 96 × 2 = 0 + 0.199 218 749 643 864 145 92;
  • 29) 0.199 218 749 643 864 145 92 × 2 = 0 + 0.398 437 499 287 728 291 84;
  • 30) 0.398 437 499 287 728 291 84 × 2 = 0 + 0.796 874 998 575 456 583 68;
  • 31) 0.796 874 998 575 456 583 68 × 2 = 1 + 0.593 749 997 150 913 167 36;
  • 32) 0.593 749 997 150 913 167 36 × 2 = 1 + 0.187 499 994 301 826 334 72;
  • 33) 0.187 499 994 301 826 334 72 × 2 = 0 + 0.374 999 988 603 652 669 44;
  • 34) 0.374 999 988 603 652 669 44 × 2 = 0 + 0.749 999 977 207 305 338 88;
  • 35) 0.749 999 977 207 305 338 88 × 2 = 1 + 0.499 999 954 414 610 677 76;
  • 36) 0.499 999 954 414 610 677 76 × 2 = 0 + 0.999 999 908 829 221 355 52;
  • 37) 0.999 999 908 829 221 355 52 × 2 = 1 + 0.999 999 817 658 442 711 04;
  • 38) 0.999 999 817 658 442 711 04 × 2 = 1 + 0.999 999 635 316 885 422 08;
  • 39) 0.999 999 635 316 885 422 08 × 2 = 1 + 0.999 999 270 633 770 844 16;
  • 40) 0.999 999 270 633 770 844 16 × 2 = 1 + 0.999 998 541 267 541 688 32;
  • 41) 0.999 998 541 267 541 688 32 × 2 = 1 + 0.999 997 082 535 083 376 64;
  • 42) 0.999 997 082 535 083 376 64 × 2 = 1 + 0.999 994 165 070 166 753 28;
  • 43) 0.999 994 165 070 166 753 28 × 2 = 1 + 0.999 988 330 140 333 506 56;
  • 44) 0.999 988 330 140 333 506 56 × 2 = 1 + 0.999 976 660 280 667 013 12;
  • 45) 0.999 976 660 280 667 013 12 × 2 = 1 + 0.999 953 320 561 334 026 24;
  • 46) 0.999 953 320 561 334 026 24 × 2 = 1 + 0.999 906 641 122 668 052 48;
  • 47) 0.999 906 641 122 668 052 48 × 2 = 1 + 0.999 813 282 245 336 104 96;
  • 48) 0.999 813 282 245 336 104 96 × 2 = 1 + 0.999 626 564 490 672 209 92;
  • 49) 0.999 626 564 490 672 209 92 × 2 = 1 + 0.999 253 128 981 344 419 84;
  • 50) 0.999 253 128 981 344 419 84 × 2 = 1 + 0.998 506 257 962 688 839 68;
  • 51) 0.998 506 257 962 688 839 68 × 2 = 1 + 0.997 012 515 925 377 679 36;
  • 52) 0.997 012 515 925 377 679 36 × 2 = 1 + 0.994 025 031 850 755 358 72;
  • 53) 0.994 025 031 850 755 358 72 × 2 = 1 + 0.988 050 063 701 510 717 44;
  • 54) 0.988 050 063 701 510 717 44 × 2 = 1 + 0.976 100 127 403 021 434 88;

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:


0.000 000 000 742 147 675 32(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 32(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:


0.000 000 000 742 147 675 32(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 -0.000 000 000 742 147 675 32 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