-0.000 000 000 742 147 588 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 588(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
-0.000 000 000 742 147 588(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:

|-0.000 000 000 742 147 588| = 0.000 000 000 742 147 588


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: 0.000 000 000 742 147 588.

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) 0.000 000 000 742 147 588 × 2 = 0 + 0.000 000 001 484 295 176;
  • 2) 0.000 000 001 484 295 176 × 2 = 0 + 0.000 000 002 968 590 352;
  • 3) 0.000 000 002 968 590 352 × 2 = 0 + 0.000 000 005 937 180 704;
  • 4) 0.000 000 005 937 180 704 × 2 = 0 + 0.000 000 011 874 361 408;
  • 5) 0.000 000 011 874 361 408 × 2 = 0 + 0.000 000 023 748 722 816;
  • 6) 0.000 000 023 748 722 816 × 2 = 0 + 0.000 000 047 497 445 632;
  • 7) 0.000 000 047 497 445 632 × 2 = 0 + 0.000 000 094 994 891 264;
  • 8) 0.000 000 094 994 891 264 × 2 = 0 + 0.000 000 189 989 782 528;
  • 9) 0.000 000 189 989 782 528 × 2 = 0 + 0.000 000 379 979 565 056;
  • 10) 0.000 000 379 979 565 056 × 2 = 0 + 0.000 000 759 959 130 112;
  • 11) 0.000 000 759 959 130 112 × 2 = 0 + 0.000 001 519 918 260 224;
  • 12) 0.000 001 519 918 260 224 × 2 = 0 + 0.000 003 039 836 520 448;
  • 13) 0.000 003 039 836 520 448 × 2 = 0 + 0.000 006 079 673 040 896;
  • 14) 0.000 006 079 673 040 896 × 2 = 0 + 0.000 012 159 346 081 792;
  • 15) 0.000 012 159 346 081 792 × 2 = 0 + 0.000 024 318 692 163 584;
  • 16) 0.000 024 318 692 163 584 × 2 = 0 + 0.000 048 637 384 327 168;
  • 17) 0.000 048 637 384 327 168 × 2 = 0 + 0.000 097 274 768 654 336;
  • 18) 0.000 097 274 768 654 336 × 2 = 0 + 0.000 194 549 537 308 672;
  • 19) 0.000 194 549 537 308 672 × 2 = 0 + 0.000 389 099 074 617 344;
  • 20) 0.000 389 099 074 617 344 × 2 = 0 + 0.000 778 198 149 234 688;
  • 21) 0.000 778 198 149 234 688 × 2 = 0 + 0.001 556 396 298 469 376;
  • 22) 0.001 556 396 298 469 376 × 2 = 0 + 0.003 112 792 596 938 752;
  • 23) 0.003 112 792 596 938 752 × 2 = 0 + 0.006 225 585 193 877 504;
  • 24) 0.006 225 585 193 877 504 × 2 = 0 + 0.012 451 170 387 755 008;
  • 25) 0.012 451 170 387 755 008 × 2 = 0 + 0.024 902 340 775 510 016;
  • 26) 0.024 902 340 775 510 016 × 2 = 0 + 0.049 804 681 551 020 032;
  • 27) 0.049 804 681 551 020 032 × 2 = 0 + 0.099 609 363 102 040 064;
  • 28) 0.099 609 363 102 040 064 × 2 = 0 + 0.199 218 726 204 080 128;
  • 29) 0.199 218 726 204 080 128 × 2 = 0 + 0.398 437 452 408 160 256;
  • 30) 0.398 437 452 408 160 256 × 2 = 0 + 0.796 874 904 816 320 512;
  • 31) 0.796 874 904 816 320 512 × 2 = 1 + 0.593 749 809 632 641 024;
  • 32) 0.593 749 809 632 641 024 × 2 = 1 + 0.187 499 619 265 282 048;
  • 33) 0.187 499 619 265 282 048 × 2 = 0 + 0.374 999 238 530 564 096;
  • 34) 0.374 999 238 530 564 096 × 2 = 0 + 0.749 998 477 061 128 192;
  • 35) 0.749 998 477 061 128 192 × 2 = 1 + 0.499 996 954 122 256 384;
  • 36) 0.499 996 954 122 256 384 × 2 = 0 + 0.999 993 908 244 512 768;
  • 37) 0.999 993 908 244 512 768 × 2 = 1 + 0.999 987 816 489 025 536;
  • 38) 0.999 987 816 489 025 536 × 2 = 1 + 0.999 975 632 978 051 072;
  • 39) 0.999 975 632 978 051 072 × 2 = 1 + 0.999 951 265 956 102 144;
  • 40) 0.999 951 265 956 102 144 × 2 = 1 + 0.999 902 531 912 204 288;
  • 41) 0.999 902 531 912 204 288 × 2 = 1 + 0.999 805 063 824 408 576;
  • 42) 0.999 805 063 824 408 576 × 2 = 1 + 0.999 610 127 648 817 152;
  • 43) 0.999 610 127 648 817 152 × 2 = 1 + 0.999 220 255 297 634 304;
  • 44) 0.999 220 255 297 634 304 × 2 = 1 + 0.998 440 510 595 268 608;
  • 45) 0.998 440 510 595 268 608 × 2 = 1 + 0.996 881 021 190 537 216;
  • 46) 0.996 881 021 190 537 216 × 2 = 1 + 0.993 762 042 381 074 432;
  • 47) 0.993 762 042 381 074 432 × 2 = 1 + 0.987 524 084 762 148 864;
  • 48) 0.987 524 084 762 148 864 × 2 = 1 + 0.975 048 169 524 297 728;
  • 49) 0.975 048 169 524 297 728 × 2 = 1 + 0.950 096 339 048 595 456;
  • 50) 0.950 096 339 048 595 456 × 2 = 1 + 0.900 192 678 097 190 912;
  • 51) 0.900 192 678 097 190 912 × 2 = 1 + 0.800 385 356 194 381 824;
  • 52) 0.800 385 356 194 381 824 × 2 = 1 + 0.600 770 712 388 763 648;
  • 53) 0.600 770 712 388 763 648 × 2 = 1 + 0.201 541 424 777 527 296;
  • 54) 0.201 541 424 777 527 296 × 2 = 0 + 0.403 082 849 555 054 592;

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:


0.000 000 000 742 147 588(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 588(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(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:


0.000 000 000 742 147 588(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2) × 20 =

1.1001 0111 1111 1111 1111 110(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 110


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 1110 =

100 1011 1111 1111 1111 1110


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 1110


Decimal number -0.000 000 000 742 147 588 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1110

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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