-0.000 000 000 742 071 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 071(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
-0.000 000 000 742 071(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:

|-0.000 000 000 742 071| = 0.000 000 000 742 071


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: 0.000 000 000 742 071.

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) 0.000 000 000 742 071 × 2 = 0 + 0.000 000 001 484 142;
  • 2) 0.000 000 001 484 142 × 2 = 0 + 0.000 000 002 968 284;
  • 3) 0.000 000 002 968 284 × 2 = 0 + 0.000 000 005 936 568;
  • 4) 0.000 000 005 936 568 × 2 = 0 + 0.000 000 011 873 136;
  • 5) 0.000 000 011 873 136 × 2 = 0 + 0.000 000 023 746 272;
  • 6) 0.000 000 023 746 272 × 2 = 0 + 0.000 000 047 492 544;
  • 7) 0.000 000 047 492 544 × 2 = 0 + 0.000 000 094 985 088;
  • 8) 0.000 000 094 985 088 × 2 = 0 + 0.000 000 189 970 176;
  • 9) 0.000 000 189 970 176 × 2 = 0 + 0.000 000 379 940 352;
  • 10) 0.000 000 379 940 352 × 2 = 0 + 0.000 000 759 880 704;
  • 11) 0.000 000 759 880 704 × 2 = 0 + 0.000 001 519 761 408;
  • 12) 0.000 001 519 761 408 × 2 = 0 + 0.000 003 039 522 816;
  • 13) 0.000 003 039 522 816 × 2 = 0 + 0.000 006 079 045 632;
  • 14) 0.000 006 079 045 632 × 2 = 0 + 0.000 012 158 091 264;
  • 15) 0.000 012 158 091 264 × 2 = 0 + 0.000 024 316 182 528;
  • 16) 0.000 024 316 182 528 × 2 = 0 + 0.000 048 632 365 056;
  • 17) 0.000 048 632 365 056 × 2 = 0 + 0.000 097 264 730 112;
  • 18) 0.000 097 264 730 112 × 2 = 0 + 0.000 194 529 460 224;
  • 19) 0.000 194 529 460 224 × 2 = 0 + 0.000 389 058 920 448;
  • 20) 0.000 389 058 920 448 × 2 = 0 + 0.000 778 117 840 896;
  • 21) 0.000 778 117 840 896 × 2 = 0 + 0.001 556 235 681 792;
  • 22) 0.001 556 235 681 792 × 2 = 0 + 0.003 112 471 363 584;
  • 23) 0.003 112 471 363 584 × 2 = 0 + 0.006 224 942 727 168;
  • 24) 0.006 224 942 727 168 × 2 = 0 + 0.012 449 885 454 336;
  • 25) 0.012 449 885 454 336 × 2 = 0 + 0.024 899 770 908 672;
  • 26) 0.024 899 770 908 672 × 2 = 0 + 0.049 799 541 817 344;
  • 27) 0.049 799 541 817 344 × 2 = 0 + 0.099 599 083 634 688;
  • 28) 0.099 599 083 634 688 × 2 = 0 + 0.199 198 167 269 376;
  • 29) 0.199 198 167 269 376 × 2 = 0 + 0.398 396 334 538 752;
  • 30) 0.398 396 334 538 752 × 2 = 0 + 0.796 792 669 077 504;
  • 31) 0.796 792 669 077 504 × 2 = 1 + 0.593 585 338 155 008;
  • 32) 0.593 585 338 155 008 × 2 = 1 + 0.187 170 676 310 016;
  • 33) 0.187 170 676 310 016 × 2 = 0 + 0.374 341 352 620 032;
  • 34) 0.374 341 352 620 032 × 2 = 0 + 0.748 682 705 240 064;
  • 35) 0.748 682 705 240 064 × 2 = 1 + 0.497 365 410 480 128;
  • 36) 0.497 365 410 480 128 × 2 = 0 + 0.994 730 820 960 256;
  • 37) 0.994 730 820 960 256 × 2 = 1 + 0.989 461 641 920 512;
  • 38) 0.989 461 641 920 512 × 2 = 1 + 0.978 923 283 841 024;
  • 39) 0.978 923 283 841 024 × 2 = 1 + 0.957 846 567 682 048;
  • 40) 0.957 846 567 682 048 × 2 = 1 + 0.915 693 135 364 096;
  • 41) 0.915 693 135 364 096 × 2 = 1 + 0.831 386 270 728 192;
  • 42) 0.831 386 270 728 192 × 2 = 1 + 0.662 772 541 456 384;
  • 43) 0.662 772 541 456 384 × 2 = 1 + 0.325 545 082 912 768;
  • 44) 0.325 545 082 912 768 × 2 = 0 + 0.651 090 165 825 536;
  • 45) 0.651 090 165 825 536 × 2 = 1 + 0.302 180 331 651 072;
  • 46) 0.302 180 331 651 072 × 2 = 0 + 0.604 360 663 302 144;
  • 47) 0.604 360 663 302 144 × 2 = 1 + 0.208 721 326 604 288;
  • 48) 0.208 721 326 604 288 × 2 = 0 + 0.417 442 653 208 576;
  • 49) 0.417 442 653 208 576 × 2 = 0 + 0.834 885 306 417 152;
  • 50) 0.834 885 306 417 152 × 2 = 1 + 0.669 770 612 834 304;
  • 51) 0.669 770 612 834 304 × 2 = 1 + 0.339 541 225 668 608;
  • 52) 0.339 541 225 668 608 × 2 = 0 + 0.679 082 451 337 216;
  • 53) 0.679 082 451 337 216 × 2 = 1 + 0.358 164 902 674 432;
  • 54) 0.358 164 902 674 432 × 2 = 0 + 0.716 329 805 348 864;

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:


0.000 000 000 742 071(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 0110 10(2)

6. Positive number before normalization:

0.000 000 000 742 071(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 0110 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:


0.000 000 000 742 071(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 0110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 0110 10(2) × 20 =

1.1001 0111 1111 0101 0011 010(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 0101 0011 010


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 1010 1001 1010 =

100 1011 1111 1010 1001 1010


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 1010 1001 1010


Decimal number -0.000 000 000 742 071 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1010 1001 1010

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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