-0.000 000 000 742 147 676 610 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 610 8(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
-0.000 000 000 742 147 676 610 8(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:

|-0.000 000 000 742 147 676 610 8| = 0.000 000 000 742 147 676 610 8


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: 0.000 000 000 742 147 676 610 8.

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) 0.000 000 000 742 147 676 610 8 × 2 = 0 + 0.000 000 001 484 295 353 221 6;
  • 2) 0.000 000 001 484 295 353 221 6 × 2 = 0 + 0.000 000 002 968 590 706 443 2;
  • 3) 0.000 000 002 968 590 706 443 2 × 2 = 0 + 0.000 000 005 937 181 412 886 4;
  • 4) 0.000 000 005 937 181 412 886 4 × 2 = 0 + 0.000 000 011 874 362 825 772 8;
  • 5) 0.000 000 011 874 362 825 772 8 × 2 = 0 + 0.000 000 023 748 725 651 545 6;
  • 6) 0.000 000 023 748 725 651 545 6 × 2 = 0 + 0.000 000 047 497 451 303 091 2;
  • 7) 0.000 000 047 497 451 303 091 2 × 2 = 0 + 0.000 000 094 994 902 606 182 4;
  • 8) 0.000 000 094 994 902 606 182 4 × 2 = 0 + 0.000 000 189 989 805 212 364 8;
  • 9) 0.000 000 189 989 805 212 364 8 × 2 = 0 + 0.000 000 379 979 610 424 729 6;
  • 10) 0.000 000 379 979 610 424 729 6 × 2 = 0 + 0.000 000 759 959 220 849 459 2;
  • 11) 0.000 000 759 959 220 849 459 2 × 2 = 0 + 0.000 001 519 918 441 698 918 4;
  • 12) 0.000 001 519 918 441 698 918 4 × 2 = 0 + 0.000 003 039 836 883 397 836 8;
  • 13) 0.000 003 039 836 883 397 836 8 × 2 = 0 + 0.000 006 079 673 766 795 673 6;
  • 14) 0.000 006 079 673 766 795 673 6 × 2 = 0 + 0.000 012 159 347 533 591 347 2;
  • 15) 0.000 012 159 347 533 591 347 2 × 2 = 0 + 0.000 024 318 695 067 182 694 4;
  • 16) 0.000 024 318 695 067 182 694 4 × 2 = 0 + 0.000 048 637 390 134 365 388 8;
  • 17) 0.000 048 637 390 134 365 388 8 × 2 = 0 + 0.000 097 274 780 268 730 777 6;
  • 18) 0.000 097 274 780 268 730 777 6 × 2 = 0 + 0.000 194 549 560 537 461 555 2;
  • 19) 0.000 194 549 560 537 461 555 2 × 2 = 0 + 0.000 389 099 121 074 923 110 4;
  • 20) 0.000 389 099 121 074 923 110 4 × 2 = 0 + 0.000 778 198 242 149 846 220 8;
  • 21) 0.000 778 198 242 149 846 220 8 × 2 = 0 + 0.001 556 396 484 299 692 441 6;
  • 22) 0.001 556 396 484 299 692 441 6 × 2 = 0 + 0.003 112 792 968 599 384 883 2;
  • 23) 0.003 112 792 968 599 384 883 2 × 2 = 0 + 0.006 225 585 937 198 769 766 4;
  • 24) 0.006 225 585 937 198 769 766 4 × 2 = 0 + 0.012 451 171 874 397 539 532 8;
  • 25) 0.012 451 171 874 397 539 532 8 × 2 = 0 + 0.024 902 343 748 795 079 065 6;
  • 26) 0.024 902 343 748 795 079 065 6 × 2 = 0 + 0.049 804 687 497 590 158 131 2;
  • 27) 0.049 804 687 497 590 158 131 2 × 2 = 0 + 0.099 609 374 995 180 316 262 4;
  • 28) 0.099 609 374 995 180 316 262 4 × 2 = 0 + 0.199 218 749 990 360 632 524 8;
  • 29) 0.199 218 749 990 360 632 524 8 × 2 = 0 + 0.398 437 499 980 721 265 049 6;
  • 30) 0.398 437 499 980 721 265 049 6 × 2 = 0 + 0.796 874 999 961 442 530 099 2;
  • 31) 0.796 874 999 961 442 530 099 2 × 2 = 1 + 0.593 749 999 922 885 060 198 4;
  • 32) 0.593 749 999 922 885 060 198 4 × 2 = 1 + 0.187 499 999 845 770 120 396 8;
  • 33) 0.187 499 999 845 770 120 396 8 × 2 = 0 + 0.374 999 999 691 540 240 793 6;
  • 34) 0.374 999 999 691 540 240 793 6 × 2 = 0 + 0.749 999 999 383 080 481 587 2;
  • 35) 0.749 999 999 383 080 481 587 2 × 2 = 1 + 0.499 999 998 766 160 963 174 4;
  • 36) 0.499 999 998 766 160 963 174 4 × 2 = 0 + 0.999 999 997 532 321 926 348 8;
  • 37) 0.999 999 997 532 321 926 348 8 × 2 = 1 + 0.999 999 995 064 643 852 697 6;
  • 38) 0.999 999 995 064 643 852 697 6 × 2 = 1 + 0.999 999 990 129 287 705 395 2;
  • 39) 0.999 999 990 129 287 705 395 2 × 2 = 1 + 0.999 999 980 258 575 410 790 4;
  • 40) 0.999 999 980 258 575 410 790 4 × 2 = 1 + 0.999 999 960 517 150 821 580 8;
  • 41) 0.999 999 960 517 150 821 580 8 × 2 = 1 + 0.999 999 921 034 301 643 161 6;
  • 42) 0.999 999 921 034 301 643 161 6 × 2 = 1 + 0.999 999 842 068 603 286 323 2;
  • 43) 0.999 999 842 068 603 286 323 2 × 2 = 1 + 0.999 999 684 137 206 572 646 4;
  • 44) 0.999 999 684 137 206 572 646 4 × 2 = 1 + 0.999 999 368 274 413 145 292 8;
  • 45) 0.999 999 368 274 413 145 292 8 × 2 = 1 + 0.999 998 736 548 826 290 585 6;
  • 46) 0.999 998 736 548 826 290 585 6 × 2 = 1 + 0.999 997 473 097 652 581 171 2;
  • 47) 0.999 997 473 097 652 581 171 2 × 2 = 1 + 0.999 994 946 195 305 162 342 4;
  • 48) 0.999 994 946 195 305 162 342 4 × 2 = 1 + 0.999 989 892 390 610 324 684 8;
  • 49) 0.999 989 892 390 610 324 684 8 × 2 = 1 + 0.999 979 784 781 220 649 369 6;
  • 50) 0.999 979 784 781 220 649 369 6 × 2 = 1 + 0.999 959 569 562 441 298 739 2;
  • 51) 0.999 959 569 562 441 298 739 2 × 2 = 1 + 0.999 919 139 124 882 597 478 4;
  • 52) 0.999 919 139 124 882 597 478 4 × 2 = 1 + 0.999 838 278 249 765 194 956 8;
  • 53) 0.999 838 278 249 765 194 956 8 × 2 = 1 + 0.999 676 556 499 530 389 913 6;
  • 54) 0.999 676 556 499 530 389 913 6 × 2 = 1 + 0.999 353 112 999 060 779 827 2;

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:


0.000 000 000 742 147 676 610 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 610 8(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:


0.000 000 000 742 147 676 610 8(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 -0.000 000 000 742 147 676 610 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 609 7 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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