-0.000 000 000 742 147 667 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 667(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
-0.000 000 000 742 147 667(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:

|-0.000 000 000 742 147 667| = 0.000 000 000 742 147 667


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: 0.000 000 000 742 147 667.

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) 0.000 000 000 742 147 667 × 2 = 0 + 0.000 000 001 484 295 334;
  • 2) 0.000 000 001 484 295 334 × 2 = 0 + 0.000 000 002 968 590 668;
  • 3) 0.000 000 002 968 590 668 × 2 = 0 + 0.000 000 005 937 181 336;
  • 4) 0.000 000 005 937 181 336 × 2 = 0 + 0.000 000 011 874 362 672;
  • 5) 0.000 000 011 874 362 672 × 2 = 0 + 0.000 000 023 748 725 344;
  • 6) 0.000 000 023 748 725 344 × 2 = 0 + 0.000 000 047 497 450 688;
  • 7) 0.000 000 047 497 450 688 × 2 = 0 + 0.000 000 094 994 901 376;
  • 8) 0.000 000 094 994 901 376 × 2 = 0 + 0.000 000 189 989 802 752;
  • 9) 0.000 000 189 989 802 752 × 2 = 0 + 0.000 000 379 979 605 504;
  • 10) 0.000 000 379 979 605 504 × 2 = 0 + 0.000 000 759 959 211 008;
  • 11) 0.000 000 759 959 211 008 × 2 = 0 + 0.000 001 519 918 422 016;
  • 12) 0.000 001 519 918 422 016 × 2 = 0 + 0.000 003 039 836 844 032;
  • 13) 0.000 003 039 836 844 032 × 2 = 0 + 0.000 006 079 673 688 064;
  • 14) 0.000 006 079 673 688 064 × 2 = 0 + 0.000 012 159 347 376 128;
  • 15) 0.000 012 159 347 376 128 × 2 = 0 + 0.000 024 318 694 752 256;
  • 16) 0.000 024 318 694 752 256 × 2 = 0 + 0.000 048 637 389 504 512;
  • 17) 0.000 048 637 389 504 512 × 2 = 0 + 0.000 097 274 779 009 024;
  • 18) 0.000 097 274 779 009 024 × 2 = 0 + 0.000 194 549 558 018 048;
  • 19) 0.000 194 549 558 018 048 × 2 = 0 + 0.000 389 099 116 036 096;
  • 20) 0.000 389 099 116 036 096 × 2 = 0 + 0.000 778 198 232 072 192;
  • 21) 0.000 778 198 232 072 192 × 2 = 0 + 0.001 556 396 464 144 384;
  • 22) 0.001 556 396 464 144 384 × 2 = 0 + 0.003 112 792 928 288 768;
  • 23) 0.003 112 792 928 288 768 × 2 = 0 + 0.006 225 585 856 577 536;
  • 24) 0.006 225 585 856 577 536 × 2 = 0 + 0.012 451 171 713 155 072;
  • 25) 0.012 451 171 713 155 072 × 2 = 0 + 0.024 902 343 426 310 144;
  • 26) 0.024 902 343 426 310 144 × 2 = 0 + 0.049 804 686 852 620 288;
  • 27) 0.049 804 686 852 620 288 × 2 = 0 + 0.099 609 373 705 240 576;
  • 28) 0.099 609 373 705 240 576 × 2 = 0 + 0.199 218 747 410 481 152;
  • 29) 0.199 218 747 410 481 152 × 2 = 0 + 0.398 437 494 820 962 304;
  • 30) 0.398 437 494 820 962 304 × 2 = 0 + 0.796 874 989 641 924 608;
  • 31) 0.796 874 989 641 924 608 × 2 = 1 + 0.593 749 979 283 849 216;
  • 32) 0.593 749 979 283 849 216 × 2 = 1 + 0.187 499 958 567 698 432;
  • 33) 0.187 499 958 567 698 432 × 2 = 0 + 0.374 999 917 135 396 864;
  • 34) 0.374 999 917 135 396 864 × 2 = 0 + 0.749 999 834 270 793 728;
  • 35) 0.749 999 834 270 793 728 × 2 = 1 + 0.499 999 668 541 587 456;
  • 36) 0.499 999 668 541 587 456 × 2 = 0 + 0.999 999 337 083 174 912;
  • 37) 0.999 999 337 083 174 912 × 2 = 1 + 0.999 998 674 166 349 824;
  • 38) 0.999 998 674 166 349 824 × 2 = 1 + 0.999 997 348 332 699 648;
  • 39) 0.999 997 348 332 699 648 × 2 = 1 + 0.999 994 696 665 399 296;
  • 40) 0.999 994 696 665 399 296 × 2 = 1 + 0.999 989 393 330 798 592;
  • 41) 0.999 989 393 330 798 592 × 2 = 1 + 0.999 978 786 661 597 184;
  • 42) 0.999 978 786 661 597 184 × 2 = 1 + 0.999 957 573 323 194 368;
  • 43) 0.999 957 573 323 194 368 × 2 = 1 + 0.999 915 146 646 388 736;
  • 44) 0.999 915 146 646 388 736 × 2 = 1 + 0.999 830 293 292 777 472;
  • 45) 0.999 830 293 292 777 472 × 2 = 1 + 0.999 660 586 585 554 944;
  • 46) 0.999 660 586 585 554 944 × 2 = 1 + 0.999 321 173 171 109 888;
  • 47) 0.999 321 173 171 109 888 × 2 = 1 + 0.998 642 346 342 219 776;
  • 48) 0.998 642 346 342 219 776 × 2 = 1 + 0.997 284 692 684 439 552;
  • 49) 0.997 284 692 684 439 552 × 2 = 1 + 0.994 569 385 368 879 104;
  • 50) 0.994 569 385 368 879 104 × 2 = 1 + 0.989 138 770 737 758 208;
  • 51) 0.989 138 770 737 758 208 × 2 = 1 + 0.978 277 541 475 516 416;
  • 52) 0.978 277 541 475 516 416 × 2 = 1 + 0.956 555 082 951 032 832;
  • 53) 0.956 555 082 951 032 832 × 2 = 1 + 0.913 110 165 902 065 664;
  • 54) 0.913 110 165 902 065 664 × 2 = 1 + 0.826 220 331 804 131 328;

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:


0.000 000 000 742 147 667(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 667(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 667(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 111


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 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 667 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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