-0.000 000 000 742 147 676 659 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 659(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
-0.000 000 000 742 147 676 659(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:

|-0.000 000 000 742 147 676 659| = 0.000 000 000 742 147 676 659


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: 0.000 000 000 742 147 676 659.

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) 0.000 000 000 742 147 676 659 × 2 = 0 + 0.000 000 001 484 295 353 318;
  • 2) 0.000 000 001 484 295 353 318 × 2 = 0 + 0.000 000 002 968 590 706 636;
  • 3) 0.000 000 002 968 590 706 636 × 2 = 0 + 0.000 000 005 937 181 413 272;
  • 4) 0.000 000 005 937 181 413 272 × 2 = 0 + 0.000 000 011 874 362 826 544;
  • 5) 0.000 000 011 874 362 826 544 × 2 = 0 + 0.000 000 023 748 725 653 088;
  • 6) 0.000 000 023 748 725 653 088 × 2 = 0 + 0.000 000 047 497 451 306 176;
  • 7) 0.000 000 047 497 451 306 176 × 2 = 0 + 0.000 000 094 994 902 612 352;
  • 8) 0.000 000 094 994 902 612 352 × 2 = 0 + 0.000 000 189 989 805 224 704;
  • 9) 0.000 000 189 989 805 224 704 × 2 = 0 + 0.000 000 379 979 610 449 408;
  • 10) 0.000 000 379 979 610 449 408 × 2 = 0 + 0.000 000 759 959 220 898 816;
  • 11) 0.000 000 759 959 220 898 816 × 2 = 0 + 0.000 001 519 918 441 797 632;
  • 12) 0.000 001 519 918 441 797 632 × 2 = 0 + 0.000 003 039 836 883 595 264;
  • 13) 0.000 003 039 836 883 595 264 × 2 = 0 + 0.000 006 079 673 767 190 528;
  • 14) 0.000 006 079 673 767 190 528 × 2 = 0 + 0.000 012 159 347 534 381 056;
  • 15) 0.000 012 159 347 534 381 056 × 2 = 0 + 0.000 024 318 695 068 762 112;
  • 16) 0.000 024 318 695 068 762 112 × 2 = 0 + 0.000 048 637 390 137 524 224;
  • 17) 0.000 048 637 390 137 524 224 × 2 = 0 + 0.000 097 274 780 275 048 448;
  • 18) 0.000 097 274 780 275 048 448 × 2 = 0 + 0.000 194 549 560 550 096 896;
  • 19) 0.000 194 549 560 550 096 896 × 2 = 0 + 0.000 389 099 121 100 193 792;
  • 20) 0.000 389 099 121 100 193 792 × 2 = 0 + 0.000 778 198 242 200 387 584;
  • 21) 0.000 778 198 242 200 387 584 × 2 = 0 + 0.001 556 396 484 400 775 168;
  • 22) 0.001 556 396 484 400 775 168 × 2 = 0 + 0.003 112 792 968 801 550 336;
  • 23) 0.003 112 792 968 801 550 336 × 2 = 0 + 0.006 225 585 937 603 100 672;
  • 24) 0.006 225 585 937 603 100 672 × 2 = 0 + 0.012 451 171 875 206 201 344;
  • 25) 0.012 451 171 875 206 201 344 × 2 = 0 + 0.024 902 343 750 412 402 688;
  • 26) 0.024 902 343 750 412 402 688 × 2 = 0 + 0.049 804 687 500 824 805 376;
  • 27) 0.049 804 687 500 824 805 376 × 2 = 0 + 0.099 609 375 001 649 610 752;
  • 28) 0.099 609 375 001 649 610 752 × 2 = 0 + 0.199 218 750 003 299 221 504;
  • 29) 0.199 218 750 003 299 221 504 × 2 = 0 + 0.398 437 500 006 598 443 008;
  • 30) 0.398 437 500 006 598 443 008 × 2 = 0 + 0.796 875 000 013 196 886 016;
  • 31) 0.796 875 000 013 196 886 016 × 2 = 1 + 0.593 750 000 026 393 772 032;
  • 32) 0.593 750 000 026 393 772 032 × 2 = 1 + 0.187 500 000 052 787 544 064;
  • 33) 0.187 500 000 052 787 544 064 × 2 = 0 + 0.375 000 000 105 575 088 128;
  • 34) 0.375 000 000 105 575 088 128 × 2 = 0 + 0.750 000 000 211 150 176 256;
  • 35) 0.750 000 000 211 150 176 256 × 2 = 1 + 0.500 000 000 422 300 352 512;
  • 36) 0.500 000 000 422 300 352 512 × 2 = 1 + 0.000 000 000 844 600 705 024;
  • 37) 0.000 000 000 844 600 705 024 × 2 = 0 + 0.000 000 001 689 201 410 048;
  • 38) 0.000 000 001 689 201 410 048 × 2 = 0 + 0.000 000 003 378 402 820 096;
  • 39) 0.000 000 003 378 402 820 096 × 2 = 0 + 0.000 000 006 756 805 640 192;
  • 40) 0.000 000 006 756 805 640 192 × 2 = 0 + 0.000 000 013 513 611 280 384;
  • 41) 0.000 000 013 513 611 280 384 × 2 = 0 + 0.000 000 027 027 222 560 768;
  • 42) 0.000 000 027 027 222 560 768 × 2 = 0 + 0.000 000 054 054 445 121 536;
  • 43) 0.000 000 054 054 445 121 536 × 2 = 0 + 0.000 000 108 108 890 243 072;
  • 44) 0.000 000 108 108 890 243 072 × 2 = 0 + 0.000 000 216 217 780 486 144;
  • 45) 0.000 000 216 217 780 486 144 × 2 = 0 + 0.000 000 432 435 560 972 288;
  • 46) 0.000 000 432 435 560 972 288 × 2 = 0 + 0.000 000 864 871 121 944 576;
  • 47) 0.000 000 864 871 121 944 576 × 2 = 0 + 0.000 001 729 742 243 889 152;
  • 48) 0.000 001 729 742 243 889 152 × 2 = 0 + 0.000 003 459 484 487 778 304;
  • 49) 0.000 003 459 484 487 778 304 × 2 = 0 + 0.000 006 918 968 975 556 608;
  • 50) 0.000 006 918 968 975 556 608 × 2 = 0 + 0.000 013 837 937 951 113 216;
  • 51) 0.000 013 837 937 951 113 216 × 2 = 0 + 0.000 027 675 875 902 226 432;
  • 52) 0.000 027 675 875 902 226 432 × 2 = 0 + 0.000 055 351 751 804 452 864;
  • 53) 0.000 055 351 751 804 452 864 × 2 = 0 + 0.000 110 703 503 608 905 728;
  • 54) 0.000 110 703 503 608 905 728 × 2 = 0 + 0.000 221 407 007 217 811 456;

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:


0.000 000 000 742 147 676 659(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 659(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:


0.000 000 000 742 147 676 659(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 -0.000 000 000 742 147 676 659 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