-0.000 000 000 742 147 676 646 723 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 723(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
-0.000 000 000 742 147 676 646 723(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:

|-0.000 000 000 742 147 676 646 723| = 0.000 000 000 742 147 676 646 723


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: 0.000 000 000 742 147 676 646 723.

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) 0.000 000 000 742 147 676 646 723 × 2 = 0 + 0.000 000 001 484 295 353 293 446;
  • 2) 0.000 000 001 484 295 353 293 446 × 2 = 0 + 0.000 000 002 968 590 706 586 892;
  • 3) 0.000 000 002 968 590 706 586 892 × 2 = 0 + 0.000 000 005 937 181 413 173 784;
  • 4) 0.000 000 005 937 181 413 173 784 × 2 = 0 + 0.000 000 011 874 362 826 347 568;
  • 5) 0.000 000 011 874 362 826 347 568 × 2 = 0 + 0.000 000 023 748 725 652 695 136;
  • 6) 0.000 000 023 748 725 652 695 136 × 2 = 0 + 0.000 000 047 497 451 305 390 272;
  • 7) 0.000 000 047 497 451 305 390 272 × 2 = 0 + 0.000 000 094 994 902 610 780 544;
  • 8) 0.000 000 094 994 902 610 780 544 × 2 = 0 + 0.000 000 189 989 805 221 561 088;
  • 9) 0.000 000 189 989 805 221 561 088 × 2 = 0 + 0.000 000 379 979 610 443 122 176;
  • 10) 0.000 000 379 979 610 443 122 176 × 2 = 0 + 0.000 000 759 959 220 886 244 352;
  • 11) 0.000 000 759 959 220 886 244 352 × 2 = 0 + 0.000 001 519 918 441 772 488 704;
  • 12) 0.000 001 519 918 441 772 488 704 × 2 = 0 + 0.000 003 039 836 883 544 977 408;
  • 13) 0.000 003 039 836 883 544 977 408 × 2 = 0 + 0.000 006 079 673 767 089 954 816;
  • 14) 0.000 006 079 673 767 089 954 816 × 2 = 0 + 0.000 012 159 347 534 179 909 632;
  • 15) 0.000 012 159 347 534 179 909 632 × 2 = 0 + 0.000 024 318 695 068 359 819 264;
  • 16) 0.000 024 318 695 068 359 819 264 × 2 = 0 + 0.000 048 637 390 136 719 638 528;
  • 17) 0.000 048 637 390 136 719 638 528 × 2 = 0 + 0.000 097 274 780 273 439 277 056;
  • 18) 0.000 097 274 780 273 439 277 056 × 2 = 0 + 0.000 194 549 560 546 878 554 112;
  • 19) 0.000 194 549 560 546 878 554 112 × 2 = 0 + 0.000 389 099 121 093 757 108 224;
  • 20) 0.000 389 099 121 093 757 108 224 × 2 = 0 + 0.000 778 198 242 187 514 216 448;
  • 21) 0.000 778 198 242 187 514 216 448 × 2 = 0 + 0.001 556 396 484 375 028 432 896;
  • 22) 0.001 556 396 484 375 028 432 896 × 2 = 0 + 0.003 112 792 968 750 056 865 792;
  • 23) 0.003 112 792 968 750 056 865 792 × 2 = 0 + 0.006 225 585 937 500 113 731 584;
  • 24) 0.006 225 585 937 500 113 731 584 × 2 = 0 + 0.012 451 171 875 000 227 463 168;
  • 25) 0.012 451 171 875 000 227 463 168 × 2 = 0 + 0.024 902 343 750 000 454 926 336;
  • 26) 0.024 902 343 750 000 454 926 336 × 2 = 0 + 0.049 804 687 500 000 909 852 672;
  • 27) 0.049 804 687 500 000 909 852 672 × 2 = 0 + 0.099 609 375 000 001 819 705 344;
  • 28) 0.099 609 375 000 001 819 705 344 × 2 = 0 + 0.199 218 750 000 003 639 410 688;
  • 29) 0.199 218 750 000 003 639 410 688 × 2 = 0 + 0.398 437 500 000 007 278 821 376;
  • 30) 0.398 437 500 000 007 278 821 376 × 2 = 0 + 0.796 875 000 000 014 557 642 752;
  • 31) 0.796 875 000 000 014 557 642 752 × 2 = 1 + 0.593 750 000 000 029 115 285 504;
  • 32) 0.593 750 000 000 029 115 285 504 × 2 = 1 + 0.187 500 000 000 058 230 571 008;
  • 33) 0.187 500 000 000 058 230 571 008 × 2 = 0 + 0.375 000 000 000 116 461 142 016;
  • 34) 0.375 000 000 000 116 461 142 016 × 2 = 0 + 0.750 000 000 000 232 922 284 032;
  • 35) 0.750 000 000 000 232 922 284 032 × 2 = 1 + 0.500 000 000 000 465 844 568 064;
  • 36) 0.500 000 000 000 465 844 568 064 × 2 = 1 + 0.000 000 000 000 931 689 136 128;
  • 37) 0.000 000 000 000 931 689 136 128 × 2 = 0 + 0.000 000 000 001 863 378 272 256;
  • 38) 0.000 000 000 001 863 378 272 256 × 2 = 0 + 0.000 000 000 003 726 756 544 512;
  • 39) 0.000 000 000 003 726 756 544 512 × 2 = 0 + 0.000 000 000 007 453 513 089 024;
  • 40) 0.000 000 000 007 453 513 089 024 × 2 = 0 + 0.000 000 000 014 907 026 178 048;
  • 41) 0.000 000 000 014 907 026 178 048 × 2 = 0 + 0.000 000 000 029 814 052 356 096;
  • 42) 0.000 000 000 029 814 052 356 096 × 2 = 0 + 0.000 000 000 059 628 104 712 192;
  • 43) 0.000 000 000 059 628 104 712 192 × 2 = 0 + 0.000 000 000 119 256 209 424 384;
  • 44) 0.000 000 000 119 256 209 424 384 × 2 = 0 + 0.000 000 000 238 512 418 848 768;
  • 45) 0.000 000 000 238 512 418 848 768 × 2 = 0 + 0.000 000 000 477 024 837 697 536;
  • 46) 0.000 000 000 477 024 837 697 536 × 2 = 0 + 0.000 000 000 954 049 675 395 072;
  • 47) 0.000 000 000 954 049 675 395 072 × 2 = 0 + 0.000 000 001 908 099 350 790 144;
  • 48) 0.000 000 001 908 099 350 790 144 × 2 = 0 + 0.000 000 003 816 198 701 580 288;
  • 49) 0.000 000 003 816 198 701 580 288 × 2 = 0 + 0.000 000 007 632 397 403 160 576;
  • 50) 0.000 000 007 632 397 403 160 576 × 2 = 0 + 0.000 000 015 264 794 806 321 152;
  • 51) 0.000 000 015 264 794 806 321 152 × 2 = 0 + 0.000 000 030 529 589 612 642 304;
  • 52) 0.000 000 030 529 589 612 642 304 × 2 = 0 + 0.000 000 061 059 179 225 284 608;
  • 53) 0.000 000 061 059 179 225 284 608 × 2 = 0 + 0.000 000 122 118 358 450 569 216;
  • 54) 0.000 000 122 118 358 450 569 216 × 2 = 0 + 0.000 000 244 236 716 901 138 432;

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:


0.000 000 000 742 147 676 646 723(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 723(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:


0.000 000 000 742 147 676 646 723(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 -0.000 000 000 742 147 676 646 723 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