-0.000 000 000 742 147 691 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 691(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
-0.000 000 000 742 147 691(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:

|-0.000 000 000 742 147 691| = 0.000 000 000 742 147 691


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: 0.000 000 000 742 147 691.

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) 0.000 000 000 742 147 691 × 2 = 0 + 0.000 000 001 484 295 382;
  • 2) 0.000 000 001 484 295 382 × 2 = 0 + 0.000 000 002 968 590 764;
  • 3) 0.000 000 002 968 590 764 × 2 = 0 + 0.000 000 005 937 181 528;
  • 4) 0.000 000 005 937 181 528 × 2 = 0 + 0.000 000 011 874 363 056;
  • 5) 0.000 000 011 874 363 056 × 2 = 0 + 0.000 000 023 748 726 112;
  • 6) 0.000 000 023 748 726 112 × 2 = 0 + 0.000 000 047 497 452 224;
  • 7) 0.000 000 047 497 452 224 × 2 = 0 + 0.000 000 094 994 904 448;
  • 8) 0.000 000 094 994 904 448 × 2 = 0 + 0.000 000 189 989 808 896;
  • 9) 0.000 000 189 989 808 896 × 2 = 0 + 0.000 000 379 979 617 792;
  • 10) 0.000 000 379 979 617 792 × 2 = 0 + 0.000 000 759 959 235 584;
  • 11) 0.000 000 759 959 235 584 × 2 = 0 + 0.000 001 519 918 471 168;
  • 12) 0.000 001 519 918 471 168 × 2 = 0 + 0.000 003 039 836 942 336;
  • 13) 0.000 003 039 836 942 336 × 2 = 0 + 0.000 006 079 673 884 672;
  • 14) 0.000 006 079 673 884 672 × 2 = 0 + 0.000 012 159 347 769 344;
  • 15) 0.000 012 159 347 769 344 × 2 = 0 + 0.000 024 318 695 538 688;
  • 16) 0.000 024 318 695 538 688 × 2 = 0 + 0.000 048 637 391 077 376;
  • 17) 0.000 048 637 391 077 376 × 2 = 0 + 0.000 097 274 782 154 752;
  • 18) 0.000 097 274 782 154 752 × 2 = 0 + 0.000 194 549 564 309 504;
  • 19) 0.000 194 549 564 309 504 × 2 = 0 + 0.000 389 099 128 619 008;
  • 20) 0.000 389 099 128 619 008 × 2 = 0 + 0.000 778 198 257 238 016;
  • 21) 0.000 778 198 257 238 016 × 2 = 0 + 0.001 556 396 514 476 032;
  • 22) 0.001 556 396 514 476 032 × 2 = 0 + 0.003 112 793 028 952 064;
  • 23) 0.003 112 793 028 952 064 × 2 = 0 + 0.006 225 586 057 904 128;
  • 24) 0.006 225 586 057 904 128 × 2 = 0 + 0.012 451 172 115 808 256;
  • 25) 0.012 451 172 115 808 256 × 2 = 0 + 0.024 902 344 231 616 512;
  • 26) 0.024 902 344 231 616 512 × 2 = 0 + 0.049 804 688 463 233 024;
  • 27) 0.049 804 688 463 233 024 × 2 = 0 + 0.099 609 376 926 466 048;
  • 28) 0.099 609 376 926 466 048 × 2 = 0 + 0.199 218 753 852 932 096;
  • 29) 0.199 218 753 852 932 096 × 2 = 0 + 0.398 437 507 705 864 192;
  • 30) 0.398 437 507 705 864 192 × 2 = 0 + 0.796 875 015 411 728 384;
  • 31) 0.796 875 015 411 728 384 × 2 = 1 + 0.593 750 030 823 456 768;
  • 32) 0.593 750 030 823 456 768 × 2 = 1 + 0.187 500 061 646 913 536;
  • 33) 0.187 500 061 646 913 536 × 2 = 0 + 0.375 000 123 293 827 072;
  • 34) 0.375 000 123 293 827 072 × 2 = 0 + 0.750 000 246 587 654 144;
  • 35) 0.750 000 246 587 654 144 × 2 = 1 + 0.500 000 493 175 308 288;
  • 36) 0.500 000 493 175 308 288 × 2 = 1 + 0.000 000 986 350 616 576;
  • 37) 0.000 000 986 350 616 576 × 2 = 0 + 0.000 001 972 701 233 152;
  • 38) 0.000 001 972 701 233 152 × 2 = 0 + 0.000 003 945 402 466 304;
  • 39) 0.000 003 945 402 466 304 × 2 = 0 + 0.000 007 890 804 932 608;
  • 40) 0.000 007 890 804 932 608 × 2 = 0 + 0.000 015 781 609 865 216;
  • 41) 0.000 015 781 609 865 216 × 2 = 0 + 0.000 031 563 219 730 432;
  • 42) 0.000 031 563 219 730 432 × 2 = 0 + 0.000 063 126 439 460 864;
  • 43) 0.000 063 126 439 460 864 × 2 = 0 + 0.000 126 252 878 921 728;
  • 44) 0.000 126 252 878 921 728 × 2 = 0 + 0.000 252 505 757 843 456;
  • 45) 0.000 252 505 757 843 456 × 2 = 0 + 0.000 505 011 515 686 912;
  • 46) 0.000 505 011 515 686 912 × 2 = 0 + 0.001 010 023 031 373 824;
  • 47) 0.001 010 023 031 373 824 × 2 = 0 + 0.002 020 046 062 747 648;
  • 48) 0.002 020 046 062 747 648 × 2 = 0 + 0.004 040 092 125 495 296;
  • 49) 0.004 040 092 125 495 296 × 2 = 0 + 0.008 080 184 250 990 592;
  • 50) 0.008 080 184 250 990 592 × 2 = 0 + 0.016 160 368 501 981 184;
  • 51) 0.016 160 368 501 981 184 × 2 = 0 + 0.032 320 737 003 962 368;
  • 52) 0.032 320 737 003 962 368 × 2 = 0 + 0.064 641 474 007 924 736;
  • 53) 0.064 641 474 007 924 736 × 2 = 0 + 0.129 282 948 015 849 472;
  • 54) 0.129 282 948 015 849 472 × 2 = 0 + 0.258 565 896 031 698 944;

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:


0.000 000 000 742 147 691(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 691(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:


0.000 000 000 742 147 691(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 -0.000 000 000 742 147 691 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