-0.000 000 000 742 147 696 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 696(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
-0.000 000 000 742 147 696(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:

|-0.000 000 000 742 147 696| = 0.000 000 000 742 147 696


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: 0.000 000 000 742 147 696.

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) 0.000 000 000 742 147 696 × 2 = 0 + 0.000 000 001 484 295 392;
  • 2) 0.000 000 001 484 295 392 × 2 = 0 + 0.000 000 002 968 590 784;
  • 3) 0.000 000 002 968 590 784 × 2 = 0 + 0.000 000 005 937 181 568;
  • 4) 0.000 000 005 937 181 568 × 2 = 0 + 0.000 000 011 874 363 136;
  • 5) 0.000 000 011 874 363 136 × 2 = 0 + 0.000 000 023 748 726 272;
  • 6) 0.000 000 023 748 726 272 × 2 = 0 + 0.000 000 047 497 452 544;
  • 7) 0.000 000 047 497 452 544 × 2 = 0 + 0.000 000 094 994 905 088;
  • 8) 0.000 000 094 994 905 088 × 2 = 0 + 0.000 000 189 989 810 176;
  • 9) 0.000 000 189 989 810 176 × 2 = 0 + 0.000 000 379 979 620 352;
  • 10) 0.000 000 379 979 620 352 × 2 = 0 + 0.000 000 759 959 240 704;
  • 11) 0.000 000 759 959 240 704 × 2 = 0 + 0.000 001 519 918 481 408;
  • 12) 0.000 001 519 918 481 408 × 2 = 0 + 0.000 003 039 836 962 816;
  • 13) 0.000 003 039 836 962 816 × 2 = 0 + 0.000 006 079 673 925 632;
  • 14) 0.000 006 079 673 925 632 × 2 = 0 + 0.000 012 159 347 851 264;
  • 15) 0.000 012 159 347 851 264 × 2 = 0 + 0.000 024 318 695 702 528;
  • 16) 0.000 024 318 695 702 528 × 2 = 0 + 0.000 048 637 391 405 056;
  • 17) 0.000 048 637 391 405 056 × 2 = 0 + 0.000 097 274 782 810 112;
  • 18) 0.000 097 274 782 810 112 × 2 = 0 + 0.000 194 549 565 620 224;
  • 19) 0.000 194 549 565 620 224 × 2 = 0 + 0.000 389 099 131 240 448;
  • 20) 0.000 389 099 131 240 448 × 2 = 0 + 0.000 778 198 262 480 896;
  • 21) 0.000 778 198 262 480 896 × 2 = 0 + 0.001 556 396 524 961 792;
  • 22) 0.001 556 396 524 961 792 × 2 = 0 + 0.003 112 793 049 923 584;
  • 23) 0.003 112 793 049 923 584 × 2 = 0 + 0.006 225 586 099 847 168;
  • 24) 0.006 225 586 099 847 168 × 2 = 0 + 0.012 451 172 199 694 336;
  • 25) 0.012 451 172 199 694 336 × 2 = 0 + 0.024 902 344 399 388 672;
  • 26) 0.024 902 344 399 388 672 × 2 = 0 + 0.049 804 688 798 777 344;
  • 27) 0.049 804 688 798 777 344 × 2 = 0 + 0.099 609 377 597 554 688;
  • 28) 0.099 609 377 597 554 688 × 2 = 0 + 0.199 218 755 195 109 376;
  • 29) 0.199 218 755 195 109 376 × 2 = 0 + 0.398 437 510 390 218 752;
  • 30) 0.398 437 510 390 218 752 × 2 = 0 + 0.796 875 020 780 437 504;
  • 31) 0.796 875 020 780 437 504 × 2 = 1 + 0.593 750 041 560 875 008;
  • 32) 0.593 750 041 560 875 008 × 2 = 1 + 0.187 500 083 121 750 016;
  • 33) 0.187 500 083 121 750 016 × 2 = 0 + 0.375 000 166 243 500 032;
  • 34) 0.375 000 166 243 500 032 × 2 = 0 + 0.750 000 332 487 000 064;
  • 35) 0.750 000 332 487 000 064 × 2 = 1 + 0.500 000 664 974 000 128;
  • 36) 0.500 000 664 974 000 128 × 2 = 1 + 0.000 001 329 948 000 256;
  • 37) 0.000 001 329 948 000 256 × 2 = 0 + 0.000 002 659 896 000 512;
  • 38) 0.000 002 659 896 000 512 × 2 = 0 + 0.000 005 319 792 001 024;
  • 39) 0.000 005 319 792 001 024 × 2 = 0 + 0.000 010 639 584 002 048;
  • 40) 0.000 010 639 584 002 048 × 2 = 0 + 0.000 021 279 168 004 096;
  • 41) 0.000 021 279 168 004 096 × 2 = 0 + 0.000 042 558 336 008 192;
  • 42) 0.000 042 558 336 008 192 × 2 = 0 + 0.000 085 116 672 016 384;
  • 43) 0.000 085 116 672 016 384 × 2 = 0 + 0.000 170 233 344 032 768;
  • 44) 0.000 170 233 344 032 768 × 2 = 0 + 0.000 340 466 688 065 536;
  • 45) 0.000 340 466 688 065 536 × 2 = 0 + 0.000 680 933 376 131 072;
  • 46) 0.000 680 933 376 131 072 × 2 = 0 + 0.001 361 866 752 262 144;
  • 47) 0.001 361 866 752 262 144 × 2 = 0 + 0.002 723 733 504 524 288;
  • 48) 0.002 723 733 504 524 288 × 2 = 0 + 0.005 447 467 009 048 576;
  • 49) 0.005 447 467 009 048 576 × 2 = 0 + 0.010 894 934 018 097 152;
  • 50) 0.010 894 934 018 097 152 × 2 = 0 + 0.021 789 868 036 194 304;
  • 51) 0.021 789 868 036 194 304 × 2 = 0 + 0.043 579 736 072 388 608;
  • 52) 0.043 579 736 072 388 608 × 2 = 0 + 0.087 159 472 144 777 216;
  • 53) 0.087 159 472 144 777 216 × 2 = 0 + 0.174 318 944 289 554 432;
  • 54) 0.174 318 944 289 554 432 × 2 = 0 + 0.348 637 888 579 108 864;

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:


0.000 000 000 742 147 696(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 696(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:


0.000 000 000 742 147 696(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 -0.000 000 000 742 147 696 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