-0.000 000 000 742 147 676 232 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 232(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
-0.000 000 000 742 147 676 232(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:

|-0.000 000 000 742 147 676 232| = 0.000 000 000 742 147 676 232


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: 0.000 000 000 742 147 676 232.

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) 0.000 000 000 742 147 676 232 × 2 = 0 + 0.000 000 001 484 295 352 464;
  • 2) 0.000 000 001 484 295 352 464 × 2 = 0 + 0.000 000 002 968 590 704 928;
  • 3) 0.000 000 002 968 590 704 928 × 2 = 0 + 0.000 000 005 937 181 409 856;
  • 4) 0.000 000 005 937 181 409 856 × 2 = 0 + 0.000 000 011 874 362 819 712;
  • 5) 0.000 000 011 874 362 819 712 × 2 = 0 + 0.000 000 023 748 725 639 424;
  • 6) 0.000 000 023 748 725 639 424 × 2 = 0 + 0.000 000 047 497 451 278 848;
  • 7) 0.000 000 047 497 451 278 848 × 2 = 0 + 0.000 000 094 994 902 557 696;
  • 8) 0.000 000 094 994 902 557 696 × 2 = 0 + 0.000 000 189 989 805 115 392;
  • 9) 0.000 000 189 989 805 115 392 × 2 = 0 + 0.000 000 379 979 610 230 784;
  • 10) 0.000 000 379 979 610 230 784 × 2 = 0 + 0.000 000 759 959 220 461 568;
  • 11) 0.000 000 759 959 220 461 568 × 2 = 0 + 0.000 001 519 918 440 923 136;
  • 12) 0.000 001 519 918 440 923 136 × 2 = 0 + 0.000 003 039 836 881 846 272;
  • 13) 0.000 003 039 836 881 846 272 × 2 = 0 + 0.000 006 079 673 763 692 544;
  • 14) 0.000 006 079 673 763 692 544 × 2 = 0 + 0.000 012 159 347 527 385 088;
  • 15) 0.000 012 159 347 527 385 088 × 2 = 0 + 0.000 024 318 695 054 770 176;
  • 16) 0.000 024 318 695 054 770 176 × 2 = 0 + 0.000 048 637 390 109 540 352;
  • 17) 0.000 048 637 390 109 540 352 × 2 = 0 + 0.000 097 274 780 219 080 704;
  • 18) 0.000 097 274 780 219 080 704 × 2 = 0 + 0.000 194 549 560 438 161 408;
  • 19) 0.000 194 549 560 438 161 408 × 2 = 0 + 0.000 389 099 120 876 322 816;
  • 20) 0.000 389 099 120 876 322 816 × 2 = 0 + 0.000 778 198 241 752 645 632;
  • 21) 0.000 778 198 241 752 645 632 × 2 = 0 + 0.001 556 396 483 505 291 264;
  • 22) 0.001 556 396 483 505 291 264 × 2 = 0 + 0.003 112 792 967 010 582 528;
  • 23) 0.003 112 792 967 010 582 528 × 2 = 0 + 0.006 225 585 934 021 165 056;
  • 24) 0.006 225 585 934 021 165 056 × 2 = 0 + 0.012 451 171 868 042 330 112;
  • 25) 0.012 451 171 868 042 330 112 × 2 = 0 + 0.024 902 343 736 084 660 224;
  • 26) 0.024 902 343 736 084 660 224 × 2 = 0 + 0.049 804 687 472 169 320 448;
  • 27) 0.049 804 687 472 169 320 448 × 2 = 0 + 0.099 609 374 944 338 640 896;
  • 28) 0.099 609 374 944 338 640 896 × 2 = 0 + 0.199 218 749 888 677 281 792;
  • 29) 0.199 218 749 888 677 281 792 × 2 = 0 + 0.398 437 499 777 354 563 584;
  • 30) 0.398 437 499 777 354 563 584 × 2 = 0 + 0.796 874 999 554 709 127 168;
  • 31) 0.796 874 999 554 709 127 168 × 2 = 1 + 0.593 749 999 109 418 254 336;
  • 32) 0.593 749 999 109 418 254 336 × 2 = 1 + 0.187 499 998 218 836 508 672;
  • 33) 0.187 499 998 218 836 508 672 × 2 = 0 + 0.374 999 996 437 673 017 344;
  • 34) 0.374 999 996 437 673 017 344 × 2 = 0 + 0.749 999 992 875 346 034 688;
  • 35) 0.749 999 992 875 346 034 688 × 2 = 1 + 0.499 999 985 750 692 069 376;
  • 36) 0.499 999 985 750 692 069 376 × 2 = 0 + 0.999 999 971 501 384 138 752;
  • 37) 0.999 999 971 501 384 138 752 × 2 = 1 + 0.999 999 943 002 768 277 504;
  • 38) 0.999 999 943 002 768 277 504 × 2 = 1 + 0.999 999 886 005 536 555 008;
  • 39) 0.999 999 886 005 536 555 008 × 2 = 1 + 0.999 999 772 011 073 110 016;
  • 40) 0.999 999 772 011 073 110 016 × 2 = 1 + 0.999 999 544 022 146 220 032;
  • 41) 0.999 999 544 022 146 220 032 × 2 = 1 + 0.999 999 088 044 292 440 064;
  • 42) 0.999 999 088 044 292 440 064 × 2 = 1 + 0.999 998 176 088 584 880 128;
  • 43) 0.999 998 176 088 584 880 128 × 2 = 1 + 0.999 996 352 177 169 760 256;
  • 44) 0.999 996 352 177 169 760 256 × 2 = 1 + 0.999 992 704 354 339 520 512;
  • 45) 0.999 992 704 354 339 520 512 × 2 = 1 + 0.999 985 408 708 679 041 024;
  • 46) 0.999 985 408 708 679 041 024 × 2 = 1 + 0.999 970 817 417 358 082 048;
  • 47) 0.999 970 817 417 358 082 048 × 2 = 1 + 0.999 941 634 834 716 164 096;
  • 48) 0.999 941 634 834 716 164 096 × 2 = 1 + 0.999 883 269 669 432 328 192;
  • 49) 0.999 883 269 669 432 328 192 × 2 = 1 + 0.999 766 539 338 864 656 384;
  • 50) 0.999 766 539 338 864 656 384 × 2 = 1 + 0.999 533 078 677 729 312 768;
  • 51) 0.999 533 078 677 729 312 768 × 2 = 1 + 0.999 066 157 355 458 625 536;
  • 52) 0.999 066 157 355 458 625 536 × 2 = 1 + 0.998 132 314 710 917 251 072;
  • 53) 0.998 132 314 710 917 251 072 × 2 = 1 + 0.996 264 629 421 834 502 144;
  • 54) 0.996 264 629 421 834 502 144 × 2 = 1 + 0.992 529 258 843 669 004 288;

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:


0.000 000 000 742 147 676 232(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 232(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:


0.000 000 000 742 147 676 232(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 -0.000 000 000 742 147 676 232 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