-0.000 000 000 742 147 676 648 63 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 648 63(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
-0.000 000 000 742 147 676 648 63(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:

|-0.000 000 000 742 147 676 648 63| = 0.000 000 000 742 147 676 648 63


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: 0.000 000 000 742 147 676 648 63.

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) 0.000 000 000 742 147 676 648 63 × 2 = 0 + 0.000 000 001 484 295 353 297 26;
  • 2) 0.000 000 001 484 295 353 297 26 × 2 = 0 + 0.000 000 002 968 590 706 594 52;
  • 3) 0.000 000 002 968 590 706 594 52 × 2 = 0 + 0.000 000 005 937 181 413 189 04;
  • 4) 0.000 000 005 937 181 413 189 04 × 2 = 0 + 0.000 000 011 874 362 826 378 08;
  • 5) 0.000 000 011 874 362 826 378 08 × 2 = 0 + 0.000 000 023 748 725 652 756 16;
  • 6) 0.000 000 023 748 725 652 756 16 × 2 = 0 + 0.000 000 047 497 451 305 512 32;
  • 7) 0.000 000 047 497 451 305 512 32 × 2 = 0 + 0.000 000 094 994 902 611 024 64;
  • 8) 0.000 000 094 994 902 611 024 64 × 2 = 0 + 0.000 000 189 989 805 222 049 28;
  • 9) 0.000 000 189 989 805 222 049 28 × 2 = 0 + 0.000 000 379 979 610 444 098 56;
  • 10) 0.000 000 379 979 610 444 098 56 × 2 = 0 + 0.000 000 759 959 220 888 197 12;
  • 11) 0.000 000 759 959 220 888 197 12 × 2 = 0 + 0.000 001 519 918 441 776 394 24;
  • 12) 0.000 001 519 918 441 776 394 24 × 2 = 0 + 0.000 003 039 836 883 552 788 48;
  • 13) 0.000 003 039 836 883 552 788 48 × 2 = 0 + 0.000 006 079 673 767 105 576 96;
  • 14) 0.000 006 079 673 767 105 576 96 × 2 = 0 + 0.000 012 159 347 534 211 153 92;
  • 15) 0.000 012 159 347 534 211 153 92 × 2 = 0 + 0.000 024 318 695 068 422 307 84;
  • 16) 0.000 024 318 695 068 422 307 84 × 2 = 0 + 0.000 048 637 390 136 844 615 68;
  • 17) 0.000 048 637 390 136 844 615 68 × 2 = 0 + 0.000 097 274 780 273 689 231 36;
  • 18) 0.000 097 274 780 273 689 231 36 × 2 = 0 + 0.000 194 549 560 547 378 462 72;
  • 19) 0.000 194 549 560 547 378 462 72 × 2 = 0 + 0.000 389 099 121 094 756 925 44;
  • 20) 0.000 389 099 121 094 756 925 44 × 2 = 0 + 0.000 778 198 242 189 513 850 88;
  • 21) 0.000 778 198 242 189 513 850 88 × 2 = 0 + 0.001 556 396 484 379 027 701 76;
  • 22) 0.001 556 396 484 379 027 701 76 × 2 = 0 + 0.003 112 792 968 758 055 403 52;
  • 23) 0.003 112 792 968 758 055 403 52 × 2 = 0 + 0.006 225 585 937 516 110 807 04;
  • 24) 0.006 225 585 937 516 110 807 04 × 2 = 0 + 0.012 451 171 875 032 221 614 08;
  • 25) 0.012 451 171 875 032 221 614 08 × 2 = 0 + 0.024 902 343 750 064 443 228 16;
  • 26) 0.024 902 343 750 064 443 228 16 × 2 = 0 + 0.049 804 687 500 128 886 456 32;
  • 27) 0.049 804 687 500 128 886 456 32 × 2 = 0 + 0.099 609 375 000 257 772 912 64;
  • 28) 0.099 609 375 000 257 772 912 64 × 2 = 0 + 0.199 218 750 000 515 545 825 28;
  • 29) 0.199 218 750 000 515 545 825 28 × 2 = 0 + 0.398 437 500 001 031 091 650 56;
  • 30) 0.398 437 500 001 031 091 650 56 × 2 = 0 + 0.796 875 000 002 062 183 301 12;
  • 31) 0.796 875 000 002 062 183 301 12 × 2 = 1 + 0.593 750 000 004 124 366 602 24;
  • 32) 0.593 750 000 004 124 366 602 24 × 2 = 1 + 0.187 500 000 008 248 733 204 48;
  • 33) 0.187 500 000 008 248 733 204 48 × 2 = 0 + 0.375 000 000 016 497 466 408 96;
  • 34) 0.375 000 000 016 497 466 408 96 × 2 = 0 + 0.750 000 000 032 994 932 817 92;
  • 35) 0.750 000 000 032 994 932 817 92 × 2 = 1 + 0.500 000 000 065 989 865 635 84;
  • 36) 0.500 000 000 065 989 865 635 84 × 2 = 1 + 0.000 000 000 131 979 731 271 68;
  • 37) 0.000 000 000 131 979 731 271 68 × 2 = 0 + 0.000 000 000 263 959 462 543 36;
  • 38) 0.000 000 000 263 959 462 543 36 × 2 = 0 + 0.000 000 000 527 918 925 086 72;
  • 39) 0.000 000 000 527 918 925 086 72 × 2 = 0 + 0.000 000 001 055 837 850 173 44;
  • 40) 0.000 000 001 055 837 850 173 44 × 2 = 0 + 0.000 000 002 111 675 700 346 88;
  • 41) 0.000 000 002 111 675 700 346 88 × 2 = 0 + 0.000 000 004 223 351 400 693 76;
  • 42) 0.000 000 004 223 351 400 693 76 × 2 = 0 + 0.000 000 008 446 702 801 387 52;
  • 43) 0.000 000 008 446 702 801 387 52 × 2 = 0 + 0.000 000 016 893 405 602 775 04;
  • 44) 0.000 000 016 893 405 602 775 04 × 2 = 0 + 0.000 000 033 786 811 205 550 08;
  • 45) 0.000 000 033 786 811 205 550 08 × 2 = 0 + 0.000 000 067 573 622 411 100 16;
  • 46) 0.000 000 067 573 622 411 100 16 × 2 = 0 + 0.000 000 135 147 244 822 200 32;
  • 47) 0.000 000 135 147 244 822 200 32 × 2 = 0 + 0.000 000 270 294 489 644 400 64;
  • 48) 0.000 000 270 294 489 644 400 64 × 2 = 0 + 0.000 000 540 588 979 288 801 28;
  • 49) 0.000 000 540 588 979 288 801 28 × 2 = 0 + 0.000 001 081 177 958 577 602 56;
  • 50) 0.000 001 081 177 958 577 602 56 × 2 = 0 + 0.000 002 162 355 917 155 205 12;
  • 51) 0.000 002 162 355 917 155 205 12 × 2 = 0 + 0.000 004 324 711 834 310 410 24;
  • 52) 0.000 004 324 711 834 310 410 24 × 2 = 0 + 0.000 008 649 423 668 620 820 48;
  • 53) 0.000 008 649 423 668 620 820 48 × 2 = 0 + 0.000 017 298 847 337 241 640 96;
  • 54) 0.000 017 298 847 337 241 640 96 × 2 = 0 + 0.000 034 597 694 674 483 281 92;

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:


0.000 000 000 742 147 676 648 63(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 648 63(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:


0.000 000 000 742 147 676 648 63(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 -0.000 000 000 742 147 676 648 63 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