-0.000 000 000 742 147 943 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 943(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
-0.000 000 000 742 147 943(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:

|-0.000 000 000 742 147 943| = 0.000 000 000 742 147 943


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: 0.000 000 000 742 147 943.

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) 0.000 000 000 742 147 943 × 2 = 0 + 0.000 000 001 484 295 886;
  • 2) 0.000 000 001 484 295 886 × 2 = 0 + 0.000 000 002 968 591 772;
  • 3) 0.000 000 002 968 591 772 × 2 = 0 + 0.000 000 005 937 183 544;
  • 4) 0.000 000 005 937 183 544 × 2 = 0 + 0.000 000 011 874 367 088;
  • 5) 0.000 000 011 874 367 088 × 2 = 0 + 0.000 000 023 748 734 176;
  • 6) 0.000 000 023 748 734 176 × 2 = 0 + 0.000 000 047 497 468 352;
  • 7) 0.000 000 047 497 468 352 × 2 = 0 + 0.000 000 094 994 936 704;
  • 8) 0.000 000 094 994 936 704 × 2 = 0 + 0.000 000 189 989 873 408;
  • 9) 0.000 000 189 989 873 408 × 2 = 0 + 0.000 000 379 979 746 816;
  • 10) 0.000 000 379 979 746 816 × 2 = 0 + 0.000 000 759 959 493 632;
  • 11) 0.000 000 759 959 493 632 × 2 = 0 + 0.000 001 519 918 987 264;
  • 12) 0.000 001 519 918 987 264 × 2 = 0 + 0.000 003 039 837 974 528;
  • 13) 0.000 003 039 837 974 528 × 2 = 0 + 0.000 006 079 675 949 056;
  • 14) 0.000 006 079 675 949 056 × 2 = 0 + 0.000 012 159 351 898 112;
  • 15) 0.000 012 159 351 898 112 × 2 = 0 + 0.000 024 318 703 796 224;
  • 16) 0.000 024 318 703 796 224 × 2 = 0 + 0.000 048 637 407 592 448;
  • 17) 0.000 048 637 407 592 448 × 2 = 0 + 0.000 097 274 815 184 896;
  • 18) 0.000 097 274 815 184 896 × 2 = 0 + 0.000 194 549 630 369 792;
  • 19) 0.000 194 549 630 369 792 × 2 = 0 + 0.000 389 099 260 739 584;
  • 20) 0.000 389 099 260 739 584 × 2 = 0 + 0.000 778 198 521 479 168;
  • 21) 0.000 778 198 521 479 168 × 2 = 0 + 0.001 556 397 042 958 336;
  • 22) 0.001 556 397 042 958 336 × 2 = 0 + 0.003 112 794 085 916 672;
  • 23) 0.003 112 794 085 916 672 × 2 = 0 + 0.006 225 588 171 833 344;
  • 24) 0.006 225 588 171 833 344 × 2 = 0 + 0.012 451 176 343 666 688;
  • 25) 0.012 451 176 343 666 688 × 2 = 0 + 0.024 902 352 687 333 376;
  • 26) 0.024 902 352 687 333 376 × 2 = 0 + 0.049 804 705 374 666 752;
  • 27) 0.049 804 705 374 666 752 × 2 = 0 + 0.099 609 410 749 333 504;
  • 28) 0.099 609 410 749 333 504 × 2 = 0 + 0.199 218 821 498 667 008;
  • 29) 0.199 218 821 498 667 008 × 2 = 0 + 0.398 437 642 997 334 016;
  • 30) 0.398 437 642 997 334 016 × 2 = 0 + 0.796 875 285 994 668 032;
  • 31) 0.796 875 285 994 668 032 × 2 = 1 + 0.593 750 571 989 336 064;
  • 32) 0.593 750 571 989 336 064 × 2 = 1 + 0.187 501 143 978 672 128;
  • 33) 0.187 501 143 978 672 128 × 2 = 0 + 0.375 002 287 957 344 256;
  • 34) 0.375 002 287 957 344 256 × 2 = 0 + 0.750 004 575 914 688 512;
  • 35) 0.750 004 575 914 688 512 × 2 = 1 + 0.500 009 151 829 377 024;
  • 36) 0.500 009 151 829 377 024 × 2 = 1 + 0.000 018 303 658 754 048;
  • 37) 0.000 018 303 658 754 048 × 2 = 0 + 0.000 036 607 317 508 096;
  • 38) 0.000 036 607 317 508 096 × 2 = 0 + 0.000 073 214 635 016 192;
  • 39) 0.000 073 214 635 016 192 × 2 = 0 + 0.000 146 429 270 032 384;
  • 40) 0.000 146 429 270 032 384 × 2 = 0 + 0.000 292 858 540 064 768;
  • 41) 0.000 292 858 540 064 768 × 2 = 0 + 0.000 585 717 080 129 536;
  • 42) 0.000 585 717 080 129 536 × 2 = 0 + 0.001 171 434 160 259 072;
  • 43) 0.001 171 434 160 259 072 × 2 = 0 + 0.002 342 868 320 518 144;
  • 44) 0.002 342 868 320 518 144 × 2 = 0 + 0.004 685 736 641 036 288;
  • 45) 0.004 685 736 641 036 288 × 2 = 0 + 0.009 371 473 282 072 576;
  • 46) 0.009 371 473 282 072 576 × 2 = 0 + 0.018 742 946 564 145 152;
  • 47) 0.018 742 946 564 145 152 × 2 = 0 + 0.037 485 893 128 290 304;
  • 48) 0.037 485 893 128 290 304 × 2 = 0 + 0.074 971 786 256 580 608;
  • 49) 0.074 971 786 256 580 608 × 2 = 0 + 0.149 943 572 513 161 216;
  • 50) 0.149 943 572 513 161 216 × 2 = 0 + 0.299 887 145 026 322 432;
  • 51) 0.299 887 145 026 322 432 × 2 = 0 + 0.599 774 290 052 644 864;
  • 52) 0.599 774 290 052 644 864 × 2 = 1 + 0.199 548 580 105 289 728;
  • 53) 0.199 548 580 105 289 728 × 2 = 0 + 0.399 097 160 210 579 456;
  • 54) 0.399 097 160 210 579 456 × 2 = 0 + 0.798 194 320 421 158 912;

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:


0.000 000 000 742 147 943(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 943(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 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:


0.000 000 000 742 147 943(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(2) × 20 =

1.1001 1000 0000 0000 0000 100(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 100


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 0100 =

100 1100 0000 0000 0000 0100


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 0100


Decimal number -0.000 000 000 742 147 943 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0100

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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