0.000 000 000 000 000 176 184 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 176 184 5(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
0.000 000 000 000 000 176 184 5(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 176 184 5.

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) 0.000 000 000 000 000 176 184 5 × 2 = 0 + 0.000 000 000 000 000 352 369;
  • 2) 0.000 000 000 000 000 352 369 × 2 = 0 + 0.000 000 000 000 000 704 738;
  • 3) 0.000 000 000 000 000 704 738 × 2 = 0 + 0.000 000 000 000 001 409 476;
  • 4) 0.000 000 000 000 001 409 476 × 2 = 0 + 0.000 000 000 000 002 818 952;
  • 5) 0.000 000 000 000 002 818 952 × 2 = 0 + 0.000 000 000 000 005 637 904;
  • 6) 0.000 000 000 000 005 637 904 × 2 = 0 + 0.000 000 000 000 011 275 808;
  • 7) 0.000 000 000 000 011 275 808 × 2 = 0 + 0.000 000 000 000 022 551 616;
  • 8) 0.000 000 000 000 022 551 616 × 2 = 0 + 0.000 000 000 000 045 103 232;
  • 9) 0.000 000 000 000 045 103 232 × 2 = 0 + 0.000 000 000 000 090 206 464;
  • 10) 0.000 000 000 000 090 206 464 × 2 = 0 + 0.000 000 000 000 180 412 928;
  • 11) 0.000 000 000 000 180 412 928 × 2 = 0 + 0.000 000 000 000 360 825 856;
  • 12) 0.000 000 000 000 360 825 856 × 2 = 0 + 0.000 000 000 000 721 651 712;
  • 13) 0.000 000 000 000 721 651 712 × 2 = 0 + 0.000 000 000 001 443 303 424;
  • 14) 0.000 000 000 001 443 303 424 × 2 = 0 + 0.000 000 000 002 886 606 848;
  • 15) 0.000 000 000 002 886 606 848 × 2 = 0 + 0.000 000 000 005 773 213 696;
  • 16) 0.000 000 000 005 773 213 696 × 2 = 0 + 0.000 000 000 011 546 427 392;
  • 17) 0.000 000 000 011 546 427 392 × 2 = 0 + 0.000 000 000 023 092 854 784;
  • 18) 0.000 000 000 023 092 854 784 × 2 = 0 + 0.000 000 000 046 185 709 568;
  • 19) 0.000 000 000 046 185 709 568 × 2 = 0 + 0.000 000 000 092 371 419 136;
  • 20) 0.000 000 000 092 371 419 136 × 2 = 0 + 0.000 000 000 184 742 838 272;
  • 21) 0.000 000 000 184 742 838 272 × 2 = 0 + 0.000 000 000 369 485 676 544;
  • 22) 0.000 000 000 369 485 676 544 × 2 = 0 + 0.000 000 000 738 971 353 088;
  • 23) 0.000 000 000 738 971 353 088 × 2 = 0 + 0.000 000 001 477 942 706 176;
  • 24) 0.000 000 001 477 942 706 176 × 2 = 0 + 0.000 000 002 955 885 412 352;
  • 25) 0.000 000 002 955 885 412 352 × 2 = 0 + 0.000 000 005 911 770 824 704;
  • 26) 0.000 000 005 911 770 824 704 × 2 = 0 + 0.000 000 011 823 541 649 408;
  • 27) 0.000 000 011 823 541 649 408 × 2 = 0 + 0.000 000 023 647 083 298 816;
  • 28) 0.000 000 023 647 083 298 816 × 2 = 0 + 0.000 000 047 294 166 597 632;
  • 29) 0.000 000 047 294 166 597 632 × 2 = 0 + 0.000 000 094 588 333 195 264;
  • 30) 0.000 000 094 588 333 195 264 × 2 = 0 + 0.000 000 189 176 666 390 528;
  • 31) 0.000 000 189 176 666 390 528 × 2 = 0 + 0.000 000 378 353 332 781 056;
  • 32) 0.000 000 378 353 332 781 056 × 2 = 0 + 0.000 000 756 706 665 562 112;
  • 33) 0.000 000 756 706 665 562 112 × 2 = 0 + 0.000 001 513 413 331 124 224;
  • 34) 0.000 001 513 413 331 124 224 × 2 = 0 + 0.000 003 026 826 662 248 448;
  • 35) 0.000 003 026 826 662 248 448 × 2 = 0 + 0.000 006 053 653 324 496 896;
  • 36) 0.000 006 053 653 324 496 896 × 2 = 0 + 0.000 012 107 306 648 993 792;
  • 37) 0.000 012 107 306 648 993 792 × 2 = 0 + 0.000 024 214 613 297 987 584;
  • 38) 0.000 024 214 613 297 987 584 × 2 = 0 + 0.000 048 429 226 595 975 168;
  • 39) 0.000 048 429 226 595 975 168 × 2 = 0 + 0.000 096 858 453 191 950 336;
  • 40) 0.000 096 858 453 191 950 336 × 2 = 0 + 0.000 193 716 906 383 900 672;
  • 41) 0.000 193 716 906 383 900 672 × 2 = 0 + 0.000 387 433 812 767 801 344;
  • 42) 0.000 387 433 812 767 801 344 × 2 = 0 + 0.000 774 867 625 535 602 688;
  • 43) 0.000 774 867 625 535 602 688 × 2 = 0 + 0.001 549 735 251 071 205 376;
  • 44) 0.001 549 735 251 071 205 376 × 2 = 0 + 0.003 099 470 502 142 410 752;
  • 45) 0.003 099 470 502 142 410 752 × 2 = 0 + 0.006 198 941 004 284 821 504;
  • 46) 0.006 198 941 004 284 821 504 × 2 = 0 + 0.012 397 882 008 569 643 008;
  • 47) 0.012 397 882 008 569 643 008 × 2 = 0 + 0.024 795 764 017 139 286 016;
  • 48) 0.024 795 764 017 139 286 016 × 2 = 0 + 0.049 591 528 034 278 572 032;
  • 49) 0.049 591 528 034 278 572 032 × 2 = 0 + 0.099 183 056 068 557 144 064;
  • 50) 0.099 183 056 068 557 144 064 × 2 = 0 + 0.198 366 112 137 114 288 128;
  • 51) 0.198 366 112 137 114 288 128 × 2 = 0 + 0.396 732 224 274 228 576 256;
  • 52) 0.396 732 224 274 228 576 256 × 2 = 0 + 0.793 464 448 548 457 152 512;
  • 53) 0.793 464 448 548 457 152 512 × 2 = 1 + 0.586 928 897 096 914 305 024;
  • 54) 0.586 928 897 096 914 305 024 × 2 = 1 + 0.173 857 794 193 828 610 048;
  • 55) 0.173 857 794 193 828 610 048 × 2 = 0 + 0.347 715 588 387 657 220 096;
  • 56) 0.347 715 588 387 657 220 096 × 2 = 0 + 0.695 431 176 775 314 440 192;
  • 57) 0.695 431 176 775 314 440 192 × 2 = 1 + 0.390 862 353 550 628 880 384;
  • 58) 0.390 862 353 550 628 880 384 × 2 = 0 + 0.781 724 707 101 257 760 768;
  • 59) 0.781 724 707 101 257 760 768 × 2 = 1 + 0.563 449 414 202 515 521 536;
  • 60) 0.563 449 414 202 515 521 536 × 2 = 1 + 0.126 898 828 405 031 043 072;
  • 61) 0.126 898 828 405 031 043 072 × 2 = 0 + 0.253 797 656 810 062 086 144;
  • 62) 0.253 797 656 810 062 086 144 × 2 = 0 + 0.507 595 313 620 124 172 288;
  • 63) 0.507 595 313 620 124 172 288 × 2 = 1 + 0.015 190 627 240 248 344 576;
  • 64) 0.015 190 627 240 248 344 576 × 2 = 0 + 0.030 381 254 480 496 689 152;
  • 65) 0.030 381 254 480 496 689 152 × 2 = 0 + 0.060 762 508 960 993 378 304;
  • 66) 0.060 762 508 960 993 378 304 × 2 = 0 + 0.121 525 017 921 986 756 608;
  • 67) 0.121 525 017 921 986 756 608 × 2 = 0 + 0.243 050 035 843 973 513 216;
  • 68) 0.243 050 035 843 973 513 216 × 2 = 0 + 0.486 100 071 687 947 026 432;
  • 69) 0.486 100 071 687 947 026 432 × 2 = 0 + 0.972 200 143 375 894 052 864;
  • 70) 0.972 200 143 375 894 052 864 × 2 = 1 + 0.944 400 286 751 788 105 728;
  • 71) 0.944 400 286 751 788 105 728 × 2 = 1 + 0.888 800 573 503 576 211 456;
  • 72) 0.888 800 573 503 576 211 456 × 2 = 1 + 0.777 601 147 007 152 422 912;
  • 73) 0.777 601 147 007 152 422 912 × 2 = 1 + 0.555 202 294 014 304 845 824;
  • 74) 0.555 202 294 014 304 845 824 × 2 = 1 + 0.110 404 588 028 609 691 648;
  • 75) 0.110 404 588 028 609 691 648 × 2 = 0 + 0.220 809 176 057 219 383 296;
  • 76) 0.220 809 176 057 219 383 296 × 2 = 0 + 0.441 618 352 114 438 766 592;

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).


4. 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:


0.000 000 000 000 000 176 184 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 1011 0010 0000 0111 1100(2)

5. Positive number before normalization:

0.000 000 000 000 000 176 184 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 1011 0010 0000 0111 1100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 53 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 176 184 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 1011 0010 0000 0111 1100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 1011 0010 0000 0111 1100(2) × 20 =

1.1001 0110 0100 0000 1111 100(2) × 2-53


7. 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 0 (a positive number)

Exponent (unadjusted): -53

Mantissa (not normalized):
1.1001 0110 0100 0000 1111 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-53 + 2(8-1) - 1 =

(-53 + 127)(10) =

74(10)


9. 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;
  • 74 ÷ 2 = 37 + 0;
  • 37 ÷ 2 = 18 + 1;
  • 18 ÷ 2 = 9 + 0;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

74(10) =

0100 1010(2)


11. 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 0010 0000 0111 1100 =

100 1011 0010 0000 0111 1100


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0100 1010

Mantissa (23 bits) =
100 1011 0010 0000 0111 1100


Decimal number 0.000 000 000 000 000 176 184 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0100 1010 - 100 1011 0010 0000 0111 1100

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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